## Calculus with Applications (10th Edition)

$$f(x)=f(x)=-x(x-3)^{2}$$ $f(x)$ concave upward on $$(-\infty, 2 )$$. $f(x)$ concave downward on $$(2, \infty )$$ Inflection point at $$(2, -2)$$
$$f(x)=-x(x-3)^{2}$$ Rewrite the function $f(x)$ as $$f(x)=-x(x^{2}-6x+9)=-x^{3}+6x^{2}-9x$$ The first derivative is \begin{aligned} f^{\prime}(x) &=-(3)x^{2}+6(2)x-9 \\ &=-3x^{2}+12x-9 , \end{aligned} and the second derivative is \begin{aligned} f^{\prime\prime}(x) &=-3(2)x+12-0 ,\\ &=-6x+12 \end{aligned} A function $f$ is concave upward on an interval if $f^{\prime\prime}(x) \gt 0$ for all $x$ in the interval, so \begin{aligned} f^{\prime\prime}(x) =-6x+12 & \gt 0 \\ -6x & \gt -12\\ x & \lt \frac{ -12}{-6}= 2 \end{aligned} so $f(x)$ is concave upward on $(-\infty, 2 )$. Also function $f$ is concave downward on an interval if $f^{\prime\prime}(x) \lt 0$ for all $x$ in the interval, so \begin{aligned} f^{\prime\prime}(x) =-6x+12 & \lt 0 \\ -6x & \lt -12\\ x & \gt \frac{ -12}{-6}= 2 \end{aligned} so $f(x)$ is concave downward on $(2, \infty )$. Since the sign of $f^{\prime\prime}(x)$ changes from positive to negative at $2$, the graph changes from concave upward to concave downward at that point, so $(2 , f(2))$ is inflection point. Now ,we find $f(2)$as follows: \begin{aligned} f(2) &=-(2)((2)-3)^{2}\\ &=-(2)(-1)^{2}\\ &=-2 \end{aligned} Finally, we have inflection point at $(2, f(2))=(2, -2)$.