To find the intervals on which the function is concave down, we must find the intervals on which the second derivative of the function is negative.

First, we must find the first and second derivatives:

The derivatives were found using the following rule:

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

Now, we can make the intervals:

Note that at the bounds of the intervals the second derivative is neither positive nor negative.

To determine the sign of the second derivative on the intervals, simply plug in any value on the interval into the second derivative function; on the first interval, the second derivative is positive, on the second it is negative, and on the third it is positive. Thus, the function is concave down on the interval .

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 equation of the line at that point, you need two things: the slope at that point and the y adjustment of the function, in the form , where m is the slope and is the y adjustment. To get the slope, find the derivative of and plug in the desired point for , giving us an answer of for the slope.

To find the y adjustment pick a point 0 in the original function. For simplicity, let's plug in , which gives us a y of 1, so an easy point is . Next plug in those values into the equation of a line, . The new equation with all parameters plugged in is

Now you simply solve for , which is .

Final equation of the line tangent to at is

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 .

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:

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:

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 .

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.

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.

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,

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Therefore the derivative is zero at . To find when it is negative plug in test points on each of the three intervals created by these zeros.

For instance,

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Hence the function is decreasing on

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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 .

Tell whether f(x) is concave up or concave down on the interval \[1,2\]

To find concave up and concave down, we need to find the second derivative of f(x).

Let's begin by finding f'(x)

Next find f ''(x)

Now, to test for concavity, plug in the endpoints of the interval:

So, on this interval, f"(x) will always be negative. This means that our function is concave down on this interval.