Since a local maximum occurs at , this tells us two pieces of information: the derivative of must be zero at and the point must lie on the graph of this function. Hence we must solve the following two equations:

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From the first equation we get or .

To find the derivative we apply the quotient rule

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Solving , we get . Plugging this into the expression for gives .

To find the local max of this function, you have to first find the derivative so you cna test values and see what's going with the slope. To find the derivative, take your exponent and multiply it by the leading coefficient and then reduce the exponent by 1. Therefore, your derivative function is: . Then, set that equal to 0 to get your critical points. . Then, set up a number line and test the regions between each critical point. When you pick a test value, plug it into the derivative function. To the left of , the function has a positive slope. Then, after , it becomes negative. After 0, it turns positive again. A local max indicates that the slope went from positive to negative. This is occuring at .

The condition reads 'the limit of f of x as x approaches three from the left'. Note that we're not looking at x as it approached negative three. Only A and B have infinite limites at positive 3 (C and D show limits as x approaches negative 3), so the answer must be one of these. We can see that from the left, B approaches positive infinity at x=3 from the left, while A approaches negative infinity at x=3 from the left, so the correct answer is A.

To find the local maxima of the function, we must determine the x values at which the first derivative of the function changes from positive to negative.

The first derivative is equal to

and was found using the following rule:

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

Using the critical values, we can create our intervals on which we examine the sign of the first derivative:

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

Next, check the sign of the first derivative on each interval by plugging in any value on the interval into the first derivative function and checking the sign. On the first interval, the first derivative is positive, on the second, it is negative, and on the third, it is positive. Thus, a local maximum exists at the value at which the first derivative changed from positive to negative, .

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 determine the local maxima for the following function, we must determine where the first derivative of the function changes from positive to negative.

The first derivative of the function is equal to

and was found using the following rule:

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

Using the critical values, we can make the intervals on which we check the sign of the first derivative:

Note that because , it comes before 5 when making the intervals. Also, note that at the endpoints of the intervals the first derivative is neither positive nor negative.

Now, we can check the sign of the first derivative by plugging in any value on each interval into the first derivative. It may seem daunting having instead of a number, but to simplify your thinking, you can let the variable be any number less than 5. After doing this, and plugging in numbers on each interval, we find that on the first interval, the first derivative is positive, on the second it is negative, and on the third it is positive. A sign change from positive to negative occured at , so the local maximum exists here.

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

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Let's take 0 from the first interval, and 6 from the second interval.

The derivative stays positive on either side of the critical point, so this function doesn't have a maximum.

The first step into finding when the extrema of a function occurs is to take the derivative and set it equal to zero:

To solve the right side of the equation, we'll need to find its roots. we may either use the quadratic formula:

or see that the equation factors into:

Either way, the only nonnegative root is 2.

To see whether this a local minima or maxima, we'll take the second derivative of our equation and plug in this value of 2:

Since the second derivative is positive, we know that this represents a local minima.

and we need to find maxima. So lets take the first derivative by the product rule where . In this case, we can set . Next, we can find the derivatives of each . Now we assemble by the product rule above to get:

.

Now we must find critical points (where , endpoints, or undefined values).

The domain of the function is all real numbers and the endpoints are and . So now we just need to find where . This will be equal to when and where . There is no value of we can choose such that so lets focus on . This is true where . We know that , so maybe we can use this to find values for which . We know that in the second quadrant, sine is positive and cosine is negative. We can check and verify that . The next value will occur in the 4th quadrant, when sine and cosine flip signs. So we can characterize these values that satisfy the equation as where .

Now we need to determine which of these critical points are maxima. so, the function is increasing from and we can show this is not a maxima. Since we're going to be testing a lot of points, let's take the second derivative and use the second derivative test. If , we have a max (think concave up). So .

At , , as (is in the second quadrant, where only sine is positive). Therefore, is a local maximum. The next value, , is in the fourth quadrant where cosine is positive, so the second derivative should be positive, meaning it should be a local minimum. The next point, should be another maximum, but it lies outside the domain. So the last endpoint, , is the last local maximum on the interval.

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.