Let f(x)=[x^3-9x]

[] --- absolute value

a. does f'(x) exist?
b. does f'(3) exist?
c. does f'(-3) exist?
d. determine all extrema of f.

I will be happy to critique your thinking. Remember, you need to work on either side of x=3 here, and on either side of x=-3.

To determine if f'(x) exists, we need to check if the derivative of f(x) exists.

a. To find the derivative of f(x), we can use the power rule for differentiation. The power rule states that if we have a function g(x) = x^n, then the derivative of g(x) is given by g'(x) = nx^(n-1).

For f(x) = |x^3 - 9x|, we can rewrite it as f(x) = (x^3 - 9x) if x^3 - 9x ≥ 0, and f(x) = -(x^3 - 9x) if x^3 - 9x < 0.

When x^3 - 9x ≥ 0, f(x) simplifies to f(x) = x^3 - 9x. Applying the power rule, we find that f'(x) = 3x^2 - 9.

Similarly, when x^3 - 9x < 0, f(x) simplifies to f(x) = -(x^3 - 9x). Taking the derivative, we find that f'(x) = -(3x^2 - 9) = -3x^2 + 9.

Since the derivative exists for both cases, the derivative of f(x) exists for all x. Thus, f'(x) exists.

b. To determine if f'(3) exists, we need to evaluate the derivative of f(x) at x = 3.

Using the derivative we found earlier, f'(x) = 3x^2 - 9, we can substitute x = 3 into the derivative equation:

f'(3) = 3(3)^2 - 9 = 3(9) - 9 = 27 - 9 = 18.

Therefore, f'(3) exists and is equal to 18.

c. Similarly, to determine if f'(-3) exists, we substitute x = -3 into the derivative equation:

f'(-3) = 3(-3)^2 - 9 = 3(9) - 9 = 27 - 9 = 18.

Therefore, f'(-3) exists and is equal to 18.

d. To find the extrema of f, we need to set f'(x) = 0 and solve for x.

Set 3x^2 - 9 = 0:
3x^2 = 9
x^2 = 3
x = ±√3.

So, the critical points where f'(x) = 0 are x = √3 and x = -√3.

To determine if these critical points are local maxima, local minima, or neither, we can use the second derivative test. However, we need to find the second derivative of f(x) first.

Taking the derivative of f'(x) = 3x^2 - 9, we get f''(x) = 6x.

Plugging in the critical points, we have f''(√3) = 6√3 and f''(-√3) = 6(-√3).

Since both f''(√3) and f''(-√3) are non-zero, the second derivative test is inconclusive.

To determine if there are extrema at x = √3 and x = -√3, we need to examine the behavior of f(x) around these points.

For x > √3, f(x) = x^3 - 9x, and for x < -√3, f(x) = -(x^3 - 9x).

By analyzing the signs of x^3 - 9x, we observe that it is positive when x > √3 and negative when x < -√3.

Therefore, x = √3 is a local minimum and x = -√3 is a local maximum.

Hence, the extrema of f are a local minimum at x = √3 and a local maximum at x = -√3.