Question

Can anyone help me with this:

Answer 1

Sure thing. What you have here is the graph of the deriviative. The derivative represents the change in the function, so you're not seeing the function itself.
I'll send you some pointers for each part. ok?

Answer 2

In part a), you want to find points of inflection. At these points the concavity of the function changes. Convacity means whether the function's graph looks like a hill or a mound. Both positive and negative slopes (or positive or negative derivatives) can exist in either a concave (hill) or convex (mound). A point of inflection is when the two switch.

Answer 3

You will find a point of inflection only when the second derivative is 0. The second derivative is the derivative of the derivative, or the slope of the graph in the image.

Answer 4

You can see that the second derivative is 0 only when the slope of f'(x) is 0. It's difficult to calculate this analytically, but it is fairly easy to see on the graph, the only point where the slope is 0, where the function f'(x) is not going up or down, is at the bottom of the semi-circle. That will be the only point of inflection.

Answer 5

Even though the slope changes from going up to going down at x=0, that's not a point of inflection, because f'(x) does not have a well defined slope at this point (it is defined to be g(x) at x= 0, and g(x) is vertical there).

Answer 6

For part b), you're trying to find the values of f(x), not f'(x).

Answer 7

A key point to remember here is that \[\int\limits_{a}^{b}f'(x) dx = f(b) - f(a).\]

The integral will be the area under the graph of f'(x) between points a and b.

Answer 8

We know f(0) = 5, look at the problem. So to find f(-4) write \[\int\limits_{-4}^{0}f'(x) dx = f(0) - f(-4)\], and convert it to: \[-(\int\limits_{-4}^{0}f'(x) dx -f(0)) = f(-4)\]. Do the same thing to get f(4)

Answer 9

Remember, the integral is the area under these halves of the graph, and we have the defining functions for each half.

Answer 10

For c), note that the slope of f(x) is positive until x = 3 ln(5/3). We know this because f'(x) is greater than 0 for that whole domain. Because of this, we know that f(x) must be increasing throughout, but decreasing afterwards. Is this kinda making sense?

Answer 11

Wow yeah, so a. x = -2, b. -(8-2pi -5) = 2pi - 3 and wow that's an integral. Can you exmpain more? And I guess c is x = 3 ln(5/3)