## Calculus with Applications (10th Edition)

$$f(x)=x^{3}-3x^{2}-24x+5; \quad\quad[-3,6]$$ Critical numbers are: $-2$ and $4$. We conclude that, the absolute maximum, $33$, occurs when $x=-2$ and the absolute minimum, $-75$ occurs when $x=4.$
$$f(x)=x^{3}-3x^{2}-24x+5; \quad\quad[-3,6]$$ First look for critical numbers in the interval $(-3,6)$. find the points where the derivative is $0$ \begin{aligned} f^{\prime}(x) &=(3)x^{2}-3(2)x-24(1)+(0)\\ &=3x^{2}-6x-24 \end{aligned} Solve the equation $f^{\prime}(x)=0$ to get \begin{aligned} f^{\prime}(x)& =3x^{2}-6x-24=0\\ &= (3x+6)(x-4)=0\\ \Rightarrow\quad\quad\quad\quad\quad\\ & (3x+6) =0 \quad \text {or} \quad (x-4) =0 \\ \\ \Rightarrow\quad\quad\quad\quad\quad\\ & 3x =-6 \quad\text {or} \quad x =4 \\ \Rightarrow\quad\quad\quad\quad\quad\\ & x =-2 \quad \text {or} \quad x =4.\\ \end{aligned} So, critical numbers are: $-2$ and $4$. Now, evaluate the function at the critical numbers and the endpoints: $$\text{when } x=0: f(-3)=(-3)^{3}-3(-3)^{2}-24(-3)+5=23 \\ \text{when } x=1: f(-2)=(-2)^{3}-3(-2)^{2}-24(-2)+5=33 \\ \text{when } x=3: f(4)=(4)^{3}-3(4)^{2}-24(4)+5=-75\\ \text{when } x=5: f(6)=(6)^{3}-3(6)^{2}-24(6)+5=-31 \\$$ We conclude that, the absolute maximum, $33$, occurs when $x=-2$ and the absolute minimum, $-75$ occurs when $x=4.$