The function f is twice differentiable and satisfies f''(x) = 6x - 12. If f has a critical point at x = 1, what must be true about f at x = 1?

Step-by-step solution

1. 1

   Translate "critical point" into an equation. For a differentiable function, a critical point is a place where the derivative vanishes. There is no corner or vertical tangent to worry about here because $f$ is twice differentiable, so

   $f'(1)=0$

   That single fact is everything the phrase gives you.

2. 2

   Evaluate the given second derivative at that point. You are handed $f''$ directly, so no differentiation is needed:

   $f''(1)=6(1)-12=6-12=-6$

3. 3

   Interpret the sign of $f''(1)$ as concavity. A negative second derivative means the slope is decreasing, so the graph bends downward near $x=1$. Since $f''(1)=-6<0$, the curve is concave down at $x=1$.

4. 4

   Apply the Second Derivative Test. The test says that if $f'(c)=0$ and $f''(c)>0$ the point is a local minimum, while if $f'(c)=0$ and $f''(c)<0$ it is a local maximum; the test is inconclusive only when $f''(c)=0$. Here both conditions of the second case hold:

   $f'(1)=0\quad\text{and}\quad f''(1)=-6<0$

5. 5

   State the conclusion and note what is not determined. $f$ has a local maximum at $x=1$. Note that $f$ itself is only known up to two constants — antidifferentiating twice gives $f(x)=x^{3}-6x^{2}+C_1x+C_2$, and the condition $f'(1)=0$ pins $C_1=9$ but leaves $C_2$ free. The value $f(1)$ is therefore unknown; only the classification of the point is forced.