AHL 5.10—Second derivatives, testing for max and min

$f^{\prime }(x)=3x^2-6x-9$.

$f^{\prime \prime }(x)=6x-6$.

$f^{\prime \prime }(4)=6(4)-6=24-6=18$.

The correct value is $18$.

$24$ comes from forgetting to subtract the constant $-6$, only using $6x$.

$-6$ results from a sign error, treating $f^{\prime \prime }(x)=-6x+6$ incorrectly.

$30$ comes from using $f^{\prime }(4)$ instead of $f^{\prime \prime }(4)$.

$18$ is obtained by correctly differentiating twice and substituting.

The sign of $f^{\prime \prime }(x)$ determines concavity, not whether $f$ is increasing or decreasing.

$f^{\prime \prime }(x)<0$ means the gradient $f^{\prime }(x)$ is decreasing, which corresponds to concave down.

The curve bends like an arch, $\cap$, on this interval.

The graph is concave down throughout $(-2,3)$.

Concave up corresponds to $f^{\prime \prime }(x)>0$, the opposite sign.

Increasing relates to $f^{\prime }(x)>0$, a different derivative than the one given.

Decreasing relates to $f^{\prime }(x)<0$, confusing first and second derivative information.

Only concave down is directly implied by $f^{\prime \prime }(x)<0$.