Math Problem Solver

$f(x)=x^{3} - 9 x^{2} + 24 x + 7$ has a local maximum $27$ at $x = 2$.

$f(x)=x^{3} - 9 x^{2} + 24 x + 7$ has no global maxima.

$f(x)=x^{3} - 9 x^{2} + 24 x + 7$ has a local minimum $23$ at $x = 4$.

$f(x)=x^{3} - 9 x^{2} + 24 x + 7$ has no global minima.

To find local extrema of a function, use the second derivative test. A critical point of a single-variable function $f$ at a point $c$ is a local minimum if $f''(c) > 0$ and a local maximum if $f''(c) < 0$.

$f(x)=x^{3} - 9 x^{2} + 24 x + 7$ has critical points at: $$\begin{aligned}x &= 2\\x &= 4\end{aligned}$$ Use the second derivative test on each critical point to determine if any of these points are local extrema.
Take the second derivative. $$f''(x) = 6 \left(x - 3\right)$$

Evaluate at $x = 2$. $$\begin{aligned}f''(x) &= 6 \left(-3 + 2\right) \\ &= -6\end{aligned}$$$-6 < 0$ so $x=2$ is a local maximum. Evaluate the function at $x = 2$ to find the value of the maximum. $$\begin{aligned}f\left(2\right)&=7 + 24 \cdot 2 - 9 \cdot 2^{2} + 2^{3} \\ &=27\end{aligned}$$$f(x)$ has a local maximum $27$ at $x = 2$.

Evaluate at $x = 4$. $$\begin{aligned}f''(x) &= 6 \left(-3 + 4\right) \\ &= 6\end{aligned}$$$6 > 0$ so $x=4$ is a local minimum. Evaluate the function at $x = 4$ to find the value of the minimum. $$\begin{aligned}f(4)&=7 + 24 \cdot 4 - 9 \cdot 4^{2} + 4^{3} \\ &=23\end{aligned}$$$f(x)$ has a local minimum $23$ at $x = 4$.