Decreasing Intervals

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AP Calculus AB › Decreasing Intervals

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Find the intervals on which the function is decreasing:

$(0, \infty)$

$(-\infty, 0)$

$(-\infty, \infty)$

The function is never decreasing

Explanation

To find the intervals on which the function is decreasing, we must find the intervals on which the first derivative of the function is negative.

So, the first derivative of the function is equal to

and was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Next, we must find the critical values, at which the first derivative is equal to zero:

Now, we make the intervals, using c as our upper and lower bound:

$(-\infty, 0), (0, \infty)$

To determine whether the first derivative is positive or negative each interval, simply plug in any number on the interval into the first derivative function. On the first interval, the first derivative is positive, while on the second interval, the first derivative is negative. Our answer is therefore $(0, \infty)$ .

On what interval(s) is the function $g(t)=\frac{t}{(t^2+1)^2}$ decreasing?

$\left(-\infty, -\frac{1}{\sqrt{3}}\right), \left(\frac{1}{\sqrt{3}}, \infty\right)$

$\left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$

$(-\infty, -1), (1, \infty)$

$\left( \frac{1}{\sqrt{3}},\infty\right)$

Explanation

The function is decreasing when the first derivative is negative. We first find when the derivative is zero. To find the derivative, we apply the quotient rule,

$f'(x)=\frac{(t^2+1)1-t2(t^2+1)2t}{(t^2+1)^4}=\frac{(t^2+1)(1-3t^2)}{(t^2+1)^4}$ .

Therefore the derivative is zero at $t=\pm \frac{1}{\sqrt{3}}$ . To find when it is negative plug in test points on each of the three intervals created by these zeros.

For instance,

$f'(-1)=-\frac{1}{4}, f'(0)=1, f'(1)=-\frac{1}{4}$ .

Hence the function is decreasing on

$\left(-\infty, -\frac{1}{\sqrt{3}}\right), \left(\frac{1}{\sqrt{3}}, \infty\right)$ .

Is the function g(x) increasing or decreasing on the interval ?

Decreasing, because g'(x) is negative on the given interval.

Decreasing, because g'(x) is positive on the given interval.

Increasing, because g'(x) is negative on the given interval.

Increasing, because g'(x) is positive on the given interval.

Explanation

Is the function g(x) increasing or decreasing on the interval ?

To tell if a function is increasing or decreasing, we need to see if its first derivative is positive or negative. let's find g'(x)

Recall that the derivative of a polynomial can be found by taking each term, multiplying by its exponent and then decreasing the exponent by 1.

Next, we need to see if the function is positive or negative over the given interval.

Begin by finding g'(-5)

So, g'(-5) is negative, but what about g'(0)?

Also negative, so our answer is:

Decreasing, because g'(x) is negative on the interval .

Find the interval(s) on which the function is decreasing:

The function is never decreasing

$(-\infty, \infty)$

$(-\infty, 0)$

$(0, \infty)$

Explanation

To determine the intervals on which the function is decreasing, we must find the intervals on which the first derivative of the function is negative.

First, we find the first derivative:

It was found using the following rule:
$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Now, we must find the critical values - values at which the first derivative is equal to zero:

Now, using the critical values as upper and lower bounds, we create the intervals on which we determine the sign of the first derivative:

$(-\infty, 0), (0, \infty)$

On the first interval, the first derivative is always positive, and on the second interval, the first derivative is always positive. (Simply plug in any point in the interval into the first derivative function and check the sign.) Thus, the function is never decreasing.

For which values of is the function $f(x)=x^4 -\frac{4}{3}x^3 -12x^2 +3$ decreasing?

$-\infty < x < -2$ and

and $3 < x < \infty$

and $2 < x < \infty$

$-\infty < x < -3$ and

This function is never decreasing.

Explanation

To determine where the function is decreasing, differentiate it:

What we are interested in are the points where . To determine these points, factor the equation:

this has solutions at

This splits the graph into 4 regions, and we can test points in each to determine if is greater than or less than 0. If it is less than zero, the function is decreasing.

$- \infty < x < -2: f'(-3)=-72$ negative/decreasing

positive/increasing

negative/decreasing

$3 < x < \infty: f'(4)=96$ positive/increasing

Is increasing or decreasing on the given interval? How do you know?

Increasing, because is positive on the interval .

Decreasing, because is negative on the interval .

Increasing, because is positive on the interval .

Decreasing, because is negative on the interval .

There is not enough information to tell whether is increasing or decreasing on the interval .

Explanation

Recall that a function is increasing at a point if its first derivative is positive, and a function is decreasing if its first derivative is negative at that point. Therfore, we should start by finding f'(x). However, I will start by combining like terms and putting f(x) in standard form:

Next, plug in each of our endpoints to see what the sign of f'(x) is.

$f'(3)=30\cdot 3^4-48\cdot 3^3+30\cdot 3^2-8\cdot 3=1380$

So f'(x) is positive on the given interval, so we know that f(x) is increasing on the given interval.

Is increasing or decreasing on the interval ?

Decreasing, because the first derivative of is negative on the function .

Increasing, because the first derivative is positive on the interval .

Decreasing, because the first derivative is positive on the interval .

The function is neither increasing nor decreasing on the interval .

Increasing because the second derivative is positive on the interval .

Explanation

To find the an increasing or decreasing interval, we need to find out if the first derivative is positive or negative on the given interval. So, find by decreasing each exponent by one and multiplying by the original number.

Next, we can find and and see if they are positive or negative.

$f'(4)=-20\cdot4^3+12\cdot4-5=-1237$

$f'(7)=-20\cdot7^3+12\cdot7-5=-6781$

Both are negative, so the slope of the line tangent to is negative, so is decreasing.

Find the interval(s) where the function is decreasing.

$\small \left [-2, -\frac{2}{3} \right ]$

$\small \left [ -\frac{2}{3}, \frac{4}{3} \right]$

$\small \left [-\frac{4}{3}, 0 \right ]$

Explanation

To find the intervals where the function decreases, apply the first derivative test. Find the derivative, set equal to , and solve to find local extrema.

$\small f '(x) = 3x^2 + 8x + 4 = 0$

$\small (3x + 2)(x + 2) = 0$

So $\small x = -\frac{2}{3}$ or $\small x = -2$ .

Next, test points in each of the intervals delineated by the potential local extrema. For example:

$\small f '(-3) = 7$ , so the function increases to the left of $\small x = -2$ .

$\small f '(-1) = -1$ , so the function decreases on the interval $\small \left [-2, -\frac{2}{3} \right]$

$\small f '(0) = 4$ , so the function increases to the right of $\small x = -\frac{2}{3}$ .

Find the interval in which the following function is decreasing.

never

always

$(-\infty,-\frac{1}{2})$

$(-\frac{1}{2},\infty)$

Explanation

To find decreasing intervals, you must find when the first derivative is less than 0. Differentiate using the power rule:

$(x^n)'=nx^{n-1}$

Thus,

Since 2 is never negative, our first derivative is never negative. Therefore, our function is never decreasing.

Determine the intervals on which the given function is decreasing:

$(-\infty , 0)$

The function is never decreasing

$(0, \infty )$

$(-\infty, \infty)$

Explanation

To determine the intervals on which the function is decreasing, we must find the intervals on which the function's first derivative is negative. To do this, we must find the first derivative and the critical value(s) at which the first derivative is equal to zero:

The derivative was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Now, setting the first derivative equal to zero, we get

So, now we can make our intervals to be analyzed (is the first derivative positive or negative on the interval?), in which c is the upper and lower bound:

$(-\infty, 0), (0, \infty)$

Note that at c the first derivative is neither positive nor negative.

On the first interval, the first derivative is always negative, so the function is always decreasing on this interval. On the second interval, the first derivative is always positive, therefore the function is increasing on this interval.

We are concerned with the interval where the function is decreasing, so $(-\infty , 0)$ is our answer.

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