Local Minimum - AP Calculus AB

First, we must find the first derivative of f(x). This is simple:

f'(x) = 8x – 5

Now, the relative extreme is found at the point that the derivative is equal to 0. Therefore, set f'(x) = 0:

0 = 8x – 5; 8x = 5; x = 5/8

Now, plug that into our original equation:

f(5/8) = 4(5/8)2 – 5(5/8) + 20 = (100/64) – (25/8) + 20 = (100 – 200)/64 + 20 = (–100/64) + 20 = (1280 – 100)/64 = 1180/64 = 295/16

Take the derivative of the function with respect to :

Plug in the point and solve for :

Now the function is .

To find , plug into the original function and solve for :

A local minimum occurs when changes from negative to positive. The first step is find .

Next, find the critical points (when or undefined).

The final step is to test on which intervals is negative and positive to identify at which of these values the minimum occurs. Notice here that the first value of falls outside of the region specified in the problem.

The regions to be tested are and .

To test the regions, choose a value in that region and plug it into to see if it is positive or negative.

is negative in the first region and positive in the second region, so the minimum occurs at .

To find the corresponding value, plug in this value into the original function, giving us a value of .

If at some point , then might be a local maxima.

So the possible solutions are

In order to determine if these points are maxima (or minima) we need to determine if the function is concave up (minima) or concave down(maxima) at these points. To do this we need the second derivative.

Taking the second derivative of our function we get:

Plugging in the x values we want to examine we see the following.

maxima

minima

Therefore the only local maxima is at the point

First, find the derivative of the function.

.

Then we find that the points where the derivative equals 0 are at and . Then picking points on the intervals between these critical points and plugging them into the derivative. Where the results are negative, the function is decreasing. Where they are positive, the function in increasing.

Using this method we see that in the on the interval the function is increasing. On the function is decreasing. On the function is increasing. Because the function is increasing to the left of and decreasing to the right, is a local maximum. Plugging it into the function, we can find the y value.

First, find the derivative of the function.

.

Then we find that the points where the derivative equals 0 are at and . Then picking points on the intervals between these critical points and plugging them into the derivative. Where the results are negative, the function is decreasing.

Where they are positive, the function in increasing. Using this method we see that in the on the interval the function is increasing. On the function is decreasing. On the function is increasing. Because the function is decreasing to the left of and increasing to the right, is a local minimum. Plugging it into the function, we can find the y value.

To find the minimum of a function, find the first derivative. In order to find the derivative of this fuction use the power rule which states, .

Given the function, and applying the power rule we find the following derivative.

Check the -value at each endpoint and when the first derivative is zero, namely

The smallest value is .

To find the minimum of a function, find the first derivative. In order to find the derivative of this fuction use the quotient rule which states,

.

Given the function, and applying the quotient rule we find the following derivative.

when and when , which indicates that has a local minimum at .

To find the local minimum of any graph, you must first take the derivative of the graph equation, set it equal to zero and solve for . To take the derivative of this equation, we must use the power rule,

.

We also must remember that the derivative of a constant is 0. Taking the derivative of the graph equation, we obtain the slope equation . Solving for x when the equation is set to 0, we obtain .

In order for to be a local minimum, the slope must increase as it passes 2 from the left. Plugging in 1 and 3 into the slope equation, we find that the slope is in fact increase from -4 to 4, therefore is a local minimum.

Plugging back into the original graph equation to solve for , we find the coordinates of the local minimum for this graph is in fact .

Find any local maxima or minima of f(x) on the interval \[-10,10\]

To begin finding local mins and maxes we need to take the first derivative of the above function.

Local minimum and maximums occur wherever the first derivative is 0.

Find the y coordinate via:

So the first bit of our answer:

But is it a maximum or minimum?

To find that, we need to know if the function is concave up or concave down at the point.

To test concavity we need the second derivative:

The second derivative is positive everywhere, so this function is concave up everywhere, making this a local minimum.

To find the least y-value, we must first find where the minimum of the function is. This is achieved by finding the derivative and then testing values. The derivative of this function is

We then need to factor that and set it equal to 0, which gives us .

We then set each expression equal to 0 to give us our critical points.

This give us critical points at . We then set up a number line and test values on each side of the values. To the left of , you can choose and plug that into the derivative. We get a positive value. In between the two critical points, you can choose , which gives us a negative value. To the right of , you can choose , which gives a positive value. To find the minimum, you must find the point where the sign changes from negative to positive. That happens at . That is the x-value of the minimum.

To find the y-value, you must plug that x-value into the original function:

.

In order to solve this equation, we must first underestand that by taking the derivative of an equation of a graph and setting it equal to zero, we can find the values of where there are critical points. Critical points are either local maxima or minima, in order to figure out which you simply plug in numbers before and after that value of in order to see whether or not the slope increases/decreases as is approaches or leaves that value.

In order to take the derivative of equation, the power rule must be applied, . Taking the derivative of the graph equation, we find that it is .

Setting the equation equal to zero and solving for , we find that the critical points for this graph are present at . Now we must determine whether these critical points are local minima or maxima.

In order to determine whether or not they are local minima/maxima, we must determine the slope behavior of the graph around these points. If the slope is positive towards a critical point and negative away from that critical point, that critical point is a local maxima. Vice versa for local minima. is approximately 1.38. is approximately -1.38. Therefore we can plug in -2,0,2 into the slope equation in order to determine the behavior of the slope around those points.

Plugging into the derivative of the graph equation, we find that slope is positive.

Plugging into the derivative of the graph equation, we find that the slope is positive as well.

Plugging into the derivative of the graph equation, we find that the slope is positive.

Because the slope is always increasing, this means that there are no local minima in this graph.

To find the local minima of a function, we must find the point(s) at which the first derivative changes from negative to positive.

The first derivative of the function is:

and was found using the following rule:

Now, we must find the critical points, the points at which the first derivative is zero:

Now, we create intervals over which to see whether the first derivative is positive or negative:

On the first interval, the first derivative is negative, while on the second interval it is positive, indicating that the point of the change in sign is the local minimum.

To find the relative minimum, you must first find the derivative of the function so that you can find the critical points. When taking the derivative, multiply the exponent by the leading coefficient and then subtract 1 from the exponent: . Then, set that equal to 0 to get your critical points: . Then, test on either side of that point to see what the behavior of the function is. To the left of 2 (plugging in 0, for example), the value is negative. To the right, the value is positive. Since the slope of the function is going from negative to positive at x=2, there is a relative minimum at x=2.

To find the local minima of the function, we must find the x value at which the function's first derivative changes from negative to positive.

First, we must find the first derivative of the function:

The derivative was found using the following rules:

,

Now, we must find the critical values (where the first derivative is equal to zero) by setting the first derivative equal to zero:

Next, we must create the intervals in which the critical value is the upper and lower limit, respectively:

On the first interval, the first derivative is less than one, and on the second interval, the first derivative is positive (simply plug in any point on the interval into the first derivative function and check the sign). The change from negative to positive occurs at x=3, so is our answer.

To find the local minimum, first find the derivative of the function. To take the derivative, multiply the exponent by the coefficient in front of the x term and then decrease the exponent by 1. Therefore, the derivative is: . Then, set that equal to 0 to get your critical point(s): . Put that point on a number line and test values on either side. To the left of 2, the values are negative. To the right of 2, the values are positive. Therefore, the slope goes from negative to positive, indicating there is a local min at x=2.

The first step is to chop up the expression into two terms: . Now, integrate each term. Whenever there is an x on the denominator, the integral is lnx. Multiply that by the numerator of 4. The integral of 1 is x. Put those together to get your answer of .

We need to differentiate term by term, applying the power rule,

This gives us

The critical points are the points where the derivative equals 0. To find those, we can use the quadratic formula:

Any local minimum will fall at a critical point where the derivative passes from negative to positive. To check this, we check a point in each of the intervals defined by the critical points:

.

Let's take -2 from the first interval, 0 from the second interval, and 2 from the third interval.

The derivative moves from negative to positive at 1, so that is the function's only local minimum.

To find where the local minima occur for the function, we need the x-values at which the first derivative changes from negative to positive.

The first derivative is

and was found using the following rule:

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

Now, using the critical values, we make the intervals on which we see whether the first derivative is positive or negative (plug in any value on the interval into the first derivative function):

On the first interval, the first derivative is positive, on the second it is negative, and on the third it is positive. So, a local minimum occurs at

because the first derivative changed from positive to negative at that point.

To determine the local minima for the function, we must determine where the first derivative changes from negative to positive.

The first derivative of the function is equal to

and was found using the following rules:

,

Now, we must find the critical values, the values at which the first derivative equals zero:

We use the critical value to make our intervals on which we check the sign of the first derivative:

Note that at the endpoints of the intervals, the first derivative is neither positive nor negative.

Next, we plug in any value on the intervals into the first derivative and check the sign. On the first interval, the first derivative is negative, while on the second, it is positive. Thus, a local mininum exists at .