Corresponding characteristics of the graphs of ƒ, ƒ', and ƒ'' - AP Calculus AB
Card 1 of 33
h(x)=\frac{f+g}{f}$
If
and 
,
then find 
.
h(x)=\frac{f+g}{f}$
If
and
then find
Tap to reveal answer
We see the answer is 
when we use the product rule.
h(x)=\frac{f+g}{f}$
$h'(x)=\frac{(f'(x)+g'(x))(f(x))-(f(x)+g(x))(f'(x))}{f^{2}$$}
$h'(x)=\frac{(2+4)(1)-(1+3)(2)}{1^{2}$$} =-2
We see the answer is
h(x)=\frac{f+g}{f}$
$h'(x)=\frac{(f'(x)+g'(x))(f(x))-(f(x)+g(x))(f'(x))}{f^{2}$$}
$h'(x)=\frac{(2+4)(1)-(1+3)(2)}{1^{2}$$} =-2
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On what intervals is the function 
both concave up and decreasing?
On what intervals is the function
Tap to reveal answer
The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get
Solving for the zero's, we see 
hits zero at 
and 
. Constructing an interval test,
, we want to know the sign's in each of these intervals. Thus, we pick a value in each of the intervals and plug it into the derivative to see if it's negative or positive. We've chosen -5, 0, and 1 to be our three values.
Thus, we can see that the derivative is only negative on the interval 
.
Repeating the process for the second derivative,
The reader can verify that this equation hits 0 at -4/3. Thus, the intervals to test for the second derivative are
. Plugging in -2 and 0, we can see that the first interval is negative and the second is positive.
Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:
If this step is confusing, try drawing it out on a number line -- the first interval is from -3 to 1/3, the second from -4/3 to infinity. They only overlap on the smaller interval of -4/3 to 1/3.
Thus, our final answer is 
The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get
Solving for the zero's, we see
Thus, we can see that the derivative is only negative on the interval
Repeating the process for the second derivative,
The reader can verify that this equation hits 0 at -4/3. Thus, the intervals to test for the second derivative are
Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:
If this step is confusing, try drawing it out on a number line -- the first interval is from -3 to 1/3, the second from -4/3 to infinity. They only overlap on the smaller interval of -4/3 to 1/3.
Thus, our final answer is
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On a closed interval, the function 
is decreasing. What can we say about 
and 
on these intervals?
On a closed interval, the function
Tap to reveal answer
If 
is decreasing, then its derivative is negative. The derivative of 
is 
, so this is telling us that 
is negative.
For 
to be decreasing, 
would have to be negative, which we don't know.
being negative has nothing to do with its slope.
For 
to be decreasing, its derivative 
would need to be negative, or, alternatively 
would have to be concave down, which we don't know.
Thus, the only correct answer is that 
is negative.
If
For
For
Thus, the only correct answer is that
← Didn't Know|Knew It →
h(x)=\frac{f+g}{f}$
If
and ,
then find .
h(x)=\frac{f+g}{f}$
If
and
then find
Tap to reveal answer
We see the answer is when we use the product rule.
h(x)=\frac{f+g}{f}$
$h'(x)=\frac{(f'(x)+g'(x))(f(x))-(f(x)+g(x))(f'(x))}{f^{2}$$}
$h'(x)=\frac{(2+4)(1)-(1+3)(2)}{1^{2}$$} =-2
We see the answer is
h(x)=\frac{f+g}{f}$
$h'(x)=\frac{(f'(x)+g'(x))(f(x))-(f(x)+g(x))(f'(x))}{f^{2}$$}
$h'(x)=\frac{(2+4)(1)-(1+3)(2)}{1^{2}$$} =-2
← Didn't Know|Knew It →
On what intervals is the function both concave up and decreasing?
On what intervals is the function
Tap to reveal answer
The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get
Solving for the zero's, we see hits zero at and . Constructing an interval test,
, we want to know the sign's in each of these intervals. Thus, we pick a value in each of the intervals and plug it into the derivative to see if it's negative or positive. We've chosen -5, 0, and 1 to be our three values.
Thus, we can see that the derivative is only negative on the interval .
Repeating the process for the second derivative,
The reader can verify that this equation hits 0 at -4/3. Thus, the intervals to test for the second derivative are
. Plugging in -2 and 0, we can see that the first interval is negative and the second is positive.
Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:
If this step is confusing, try drawing it out on a number line -- the first interval is from -3 to 1/3, the second from -4/3 to infinity. They only overlap on the smaller interval of -4/3 to 1/3.
Thus, our final answer is
The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get
Solving for the zero's, we see
Thus, we can see that the derivative is only negative on the interval
Repeating the process for the second derivative,
The reader can verify that this equation hits 0 at -4/3. Thus, the intervals to test for the second derivative are
Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:
If this step is confusing, try drawing it out on a number line -- the first interval is from -3 to 1/3, the second from -4/3 to infinity. They only overlap on the smaller interval of -4/3 to 1/3.
Thus, our final answer is
← Didn't Know|Knew It →
On a closed interval, the function is decreasing. What can we say about and on these intervals?
On a closed interval, the function
Tap to reveal answer
If is decreasing, then its derivative is negative. The derivative of is , so this is telling us that is negative.
For to be decreasing, would have to be negative, which we don't know.
being negative has nothing to do with its slope.
For to be decreasing, its derivative would need to be negative, or, alternatively would have to be concave down, which we don't know.
Thus, the only correct answer is that is negative.
If
For
For
Thus, the only correct answer is that
← Didn't Know|Knew It →
h(x)=\frac{f+g}{f}$
If
and ,
then find .
h(x)=\frac{f+g}{f}$
If
and
then find
Tap to reveal answer
We see the answer is when we use the product rule.
h(x)=\frac{f+g}{f}$
$h'(x)=\frac{(f'(x)+g'(x))(f(x))-(f(x)+g(x))(f'(x))}{f^{2}$$}
$h'(x)=\frac{(2+4)(1)-(1+3)(2)}{1^{2}$$} =-2
We see the answer is
h(x)=\frac{f+g}{f}$
$h'(x)=\frac{(f'(x)+g'(x))(f(x))-(f(x)+g(x))(f'(x))}{f^{2}$$}
$h'(x)=\frac{(2+4)(1)-(1+3)(2)}{1^{2}$$} =-2
← Didn't Know|Knew It →
On what intervals is the function both concave up and decreasing?
On what intervals is the function
Tap to reveal answer
The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get
Solving for the zero's, we see hits zero at and . Constructing an interval test,
, we want to know the sign's in each of these intervals. Thus, we pick a value in each of the intervals and plug it into the derivative to see if it's negative or positive. We've chosen -5, 0, and 1 to be our three values.
Thus, we can see that the derivative is only negative on the interval .
Repeating the process for the second derivative,
The reader can verify that this equation hits 0 at -4/3. Thus, the intervals to test for the second derivative are
. Plugging in -2 and 0, we can see that the first interval is negative and the second is positive.
Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:
If this step is confusing, try drawing it out on a number line -- the first interval is from -3 to 1/3, the second from -4/3 to infinity. They only overlap on the smaller interval of -4/3 to 1/3.
Thus, our final answer is
The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get
Solving for the zero's, we see
Thus, we can see that the derivative is only negative on the interval
Repeating the process for the second derivative,
The reader can verify that this equation hits 0 at -4/3. Thus, the intervals to test for the second derivative are
Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:
If this step is confusing, try drawing it out on a number line -- the first interval is from -3 to 1/3, the second from -4/3 to infinity. They only overlap on the smaller interval of -4/3 to 1/3.
Thus, our final answer is
← Didn't Know|Knew It →
On a closed interval, the function is decreasing. What can we say about and on these intervals?
On a closed interval, the function
Tap to reveal answer
If is decreasing, then its derivative is negative. The derivative of is , so this is telling us that is negative.
For to be decreasing, would have to be negative, which we don't know.
being negative has nothing to do with its slope.
For to be decreasing, its derivative would need to be negative, or, alternatively would have to be concave down, which we don't know.
Thus, the only correct answer is that is negative.
If
For
For
Thus, the only correct answer is that
← Didn't Know|Knew It →
h(x)=\frac{f+g}{f}$
If
and ,
then find .
h(x)=\frac{f+g}{f}$
If
and
then find
Tap to reveal answer
We see the answer is when we use the product rule.
h(x)=\frac{f+g}{f}$
$h'(x)=\frac{(f'(x)+g'(x))(f(x))-(f(x)+g(x))(f'(x))}{f^{2}$$}
$h'(x)=\frac{(2+4)(1)-(1+3)(2)}{1^{2}$$} =-2
We see the answer is
h(x)=\frac{f+g}{f}$
$h'(x)=\frac{(f'(x)+g'(x))(f(x))-(f(x)+g(x))(f'(x))}{f^{2}$$}
$h'(x)=\frac{(2+4)(1)-(1+3)(2)}{1^{2}$$} =-2
← Didn't Know|Knew It →
On what intervals is the function both concave up and decreasing?
On what intervals is the function
Tap to reveal answer
The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get
Solving for the zero's, we see hits zero at and . Constructing an interval test,
, we want to know the sign's in each of these intervals. Thus, we pick a value in each of the intervals and plug it into the derivative to see if it's negative or positive. We've chosen -5, 0, and 1 to be our three values.
Thus, we can see that the derivative is only negative on the interval .
Repeating the process for the second derivative,
The reader can verify that this equation hits 0 at -4/3. Thus, the intervals to test for the second derivative are
. Plugging in -2 and 0, we can see that the first interval is negative and the second is positive.
Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:
If this step is confusing, try drawing it out on a number line -- the first interval is from -3 to 1/3, the second from -4/3 to infinity. They only overlap on the smaller interval of -4/3 to 1/3.
Thus, our final answer is
The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get
Solving for the zero's, we see
Thus, we can see that the derivative is only negative on the interval
Repeating the process for the second derivative,
The reader can verify that this equation hits 0 at -4/3. Thus, the intervals to test for the second derivative are
Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:
If this step is confusing, try drawing it out on a number line -- the first interval is from -3 to 1/3, the second from -4/3 to infinity. They only overlap on the smaller interval of -4/3 to 1/3.
Thus, our final answer is
← Didn't Know|Knew It →
On a closed interval, the function is decreasing. What can we say about and on these intervals?
On a closed interval, the function
Tap to reveal answer
If is decreasing, then its derivative is negative. The derivative of is , so this is telling us that is negative.
For to be decreasing, would have to be negative, which we don't know.
being negative has nothing to do with its slope.
For to be decreasing, its derivative would need to be negative, or, alternatively would have to be concave down, which we don't know.
Thus, the only correct answer is that is negative.
If
For
For
Thus, the only correct answer is that
← Didn't Know|Knew It →
h(x)=\frac{f+g}{f}$
If
and ,
then find .
h(x)=\frac{f+g}{f}$
If
and
then find
Tap to reveal answer
We see the answer is when we use the product rule.
h(x)=\frac{f+g}{f}$
$h'(x)=\frac{(f'(x)+g'(x))(f(x))-(f(x)+g(x))(f'(x))}{f^{2}$$}
$h'(x)=\frac{(2+4)(1)-(1+3)(2)}{1^{2}$$} =-2
We see the answer is
h(x)=\frac{f+g}{f}$
$h'(x)=\frac{(f'(x)+g'(x))(f(x))-(f(x)+g(x))(f'(x))}{f^{2}$$}
$h'(x)=\frac{(2+4)(1)-(1+3)(2)}{1^{2}$$} =-2
← Didn't Know|Knew It →
On what intervals is the function both concave up and decreasing?
On what intervals is the function
Tap to reveal answer
The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get
Solving for the zero's, we see hits zero at and . Constructing an interval test,
, we want to know the sign's in each of these intervals. Thus, we pick a value in each of the intervals and plug it into the derivative to see if it's negative or positive. We've chosen -5, 0, and 1 to be our three values.
Thus, we can see that the derivative is only negative on the interval .
Repeating the process for the second derivative,
The reader can verify that this equation hits 0 at -4/3. Thus, the intervals to test for the second derivative are
. Plugging in -2 and 0, we can see that the first interval is negative and the second is positive.
Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:
If this step is confusing, try drawing it out on a number line -- the first interval is from -3 to 1/3, the second from -4/3 to infinity. They only overlap on the smaller interval of -4/3 to 1/3.
Thus, our final answer is
The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get
Solving for the zero's, we see
Thus, we can see that the derivative is only negative on the interval
Repeating the process for the second derivative,
The reader can verify that this equation hits 0 at -4/3. Thus, the intervals to test for the second derivative are
Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:
If this step is confusing, try drawing it out on a number line -- the first interval is from -3 to 1/3, the second from -4/3 to infinity. They only overlap on the smaller interval of -4/3 to 1/3.
Thus, our final answer is
← Didn't Know|Knew It →
On a closed interval, the function is decreasing. What can we say about and on these intervals?
On a closed interval, the function
Tap to reveal answer
If is decreasing, then its derivative is negative. The derivative of is , so this is telling us that is negative.
For to be decreasing, would have to be negative, which we don't know.
being negative has nothing to do with its slope.
For to be decreasing, its derivative would need to be negative, or, alternatively would have to be concave down, which we don't know.
Thus, the only correct answer is that is negative.
If
For
For
Thus, the only correct answer is that
← Didn't Know|Knew It →
h(x)=\frac{f+g}{f}$
If
and ,
then find .
h(x)=\frac{f+g}{f}$
If
and
then find
Tap to reveal answer
We see the answer is when we use the product rule.
h(x)=\frac{f+g}{f}$
$h'(x)=\frac{(f'(x)+g'(x))(f(x))-(f(x)+g(x))(f'(x))}{f^{2}$$}
$h'(x)=\frac{(2+4)(1)-(1+3)(2)}{1^{2}$$} =-2
We see the answer is
h(x)=\frac{f+g}{f}$
$h'(x)=\frac{(f'(x)+g'(x))(f(x))-(f(x)+g(x))(f'(x))}{f^{2}$$}
$h'(x)=\frac{(2+4)(1)-(1+3)(2)}{1^{2}$$} =-2
← Didn't Know|Knew It →
On what intervals is the function both concave up and decreasing?
On what intervals is the function
Tap to reveal answer
The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get
Solving for the zero's, we see hits zero at and . Constructing an interval test,
, we want to know the sign's in each of these intervals. Thus, we pick a value in each of the intervals and plug it into the derivative to see if it's negative or positive. We've chosen -5, 0, and 1 to be our three values.
Thus, we can see that the derivative is only negative on the interval .
Repeating the process for the second derivative,
The reader can verify that this equation hits 0 at -4/3. Thus, the intervals to test for the second derivative are
. Plugging in -2 and 0, we can see that the first interval is negative and the second is positive.
Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:
If this step is confusing, try drawing it out on a number line -- the first interval is from -3 to 1/3, the second from -4/3 to infinity. They only overlap on the smaller interval of -4/3 to 1/3.
Thus, our final answer is
The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get
Solving for the zero's, we see
Thus, we can see that the derivative is only negative on the interval
Repeating the process for the second derivative,
The reader can verify that this equation hits 0 at -4/3. Thus, the intervals to test for the second derivative are
Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:
If this step is confusing, try drawing it out on a number line -- the first interval is from -3 to 1/3, the second from -4/3 to infinity. They only overlap on the smaller interval of -4/3 to 1/3.
Thus, our final answer is
← Didn't Know|Knew It →
On a closed interval, the function is decreasing. What can we say about and on these intervals?
On a closed interval, the function
Tap to reveal answer
If is decreasing, then its derivative is negative. The derivative of is , so this is telling us that is negative.
For to be decreasing, would have to be negative, which we don't know.
being negative has nothing to do with its slope.
For to be decreasing, its derivative would need to be negative, or, alternatively would have to be concave down, which we don't know.
Thus, the only correct answer is that is negative.
If
For
For
Thus, the only correct answer is that
← Didn't Know|Knew It →
h(x)=\frac{f+g}{f}$
If
and ,
then find .
h(x)=\frac{f+g}{f}$
If
and
then find
Tap to reveal answer
We see the answer is when we use the product rule.
h(x)=\frac{f+g}{f}$
$h'(x)=\frac{(f'(x)+g'(x))(f(x))-(f(x)+g(x))(f'(x))}{f^{2}$$}
$h'(x)=\frac{(2+4)(1)-(1+3)(2)}{1^{2}$$} =-2
We see the answer is
h(x)=\frac{f+g}{f}$
$h'(x)=\frac{(f'(x)+g'(x))(f(x))-(f(x)+g(x))(f'(x))}{f^{2}$$}
$h'(x)=\frac{(2+4)(1)-(1+3)(2)}{1^{2}$$} =-2
← Didn't Know|Knew It →
On what intervals is the function both concave up and decreasing?
On what intervals is the function
Tap to reveal answer
The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get
Solving for the zero's, we see hits zero at and . Constructing an interval test,
, we want to know the sign's in each of these intervals. Thus, we pick a value in each of the intervals and plug it into the derivative to see if it's negative or positive. We've chosen -5, 0, and 1 to be our three values.
Thus, we can see that the derivative is only negative on the interval .
Repeating the process for the second derivative,
The reader can verify that this equation hits 0 at -4/3. Thus, the intervals to test for the second derivative are
. Plugging in -2 and 0, we can see that the first interval is negative and the second is positive.
Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:
If this step is confusing, try drawing it out on a number line -- the first interval is from -3 to 1/3, the second from -4/3 to infinity. They only overlap on the smaller interval of -4/3 to 1/3.
Thus, our final answer is
The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get
Solving for the zero's, we see
Thus, we can see that the derivative is only negative on the interval
Repeating the process for the second derivative,
The reader can verify that this equation hits 0 at -4/3. Thus, the intervals to test for the second derivative are
Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:
If this step is confusing, try drawing it out on a number line -- the first interval is from -3 to 1/3, the second from -4/3 to infinity. They only overlap on the smaller interval of -4/3 to 1/3.
Thus, our final answer is
← Didn't Know|Knew It →