Are any of these statements true about f(x) = 3x4 - 8x3 - 24x2 + 96x + 48..... ?

Question

f(x) has a horizontal tangent line where x = 2
f(x) has a local maximum where x = 2
f(x) has a local minimum where x = 2
f(x) has an inflection point where x = -1

Answer 1

First, you take the first and second derivatives of f(x).

f'(x) = 12x^3 - 24x^2 - 48x + 96 = 12x^2(x-2) -48(x-2)
= 12(x^2-4)(x-2) = 12(x+2)(x-2)(x-2)

and notice that f'(x) has zeroes at x=2 and x=-2

f''(x) = 36x^2 - 48x -48 = 12(3x^2 - 4x - 4) = 12(3x-2)(x-2)
which has zeroes at 2/3 and 2.

Since f'(2) = 0, the first statement is true.
Since f''(2) = 0, then the function is neither concave up nor concave down at x=2 so this is not local max or min. It is an inflection point.

And, since f''(-1) does not = 0, the last statement is also false.

Answer 2

Tbolling2 is mostly right. The first statement is true, and the other three are false.

However, I should point out that it is possible for a function f(x) to have a second derivative f''(x)=0, for some value of x, and yet for the function to be concave up or down, and therefore have a local min or max, at that point. The most obvious case is f(x)=x^4. In this case, f''(x)=12x^2, which means that f''(x)=0 at x = 0. But, a quick look at the graph shows that x^4 is concave up (and therefore has a minimum) at x=0.

It just happens to be the case that this situation doesn't occur in the function given by J Russell.

Answer 3

You could solve this by analyzing the first and second derivatives, but it's easier just to graph it on a computer or graphing calculator

None of the statements are true.

In fact, this function has a local minimum at -2 and a point of inflection at x = 2.