## Calculus with Applications (10th Edition)

$$f(x)=x^{4}-18x^{2}+1; \quad\quad[-4,4]$$ Critical numbers are: $0$, $-3$ and $3$. The absolute maximum, $1$, occurs when $x=0$ and the absolute minimum, $-80$ occurs when $x=-3$ and $3.$
$$f(x)=x^{4}-18x^{2}+1; \quad\quad[-4,4]$$ First look for critical numbers in the interval $(-4,4)$. find the points where the derivative is $0$ \begin{aligned} f^{\prime}(x) &=(4)x^{3}-18(2)x+(0);\\ &=4x^{3}-36x \end{aligned} Solve the equation $f^{\prime}(x)=0$ to get \begin{aligned} f^{\prime}(x)& =4x^{3}-36x=0\\ &= 4x(x^{2}-9)=0\\ \Rightarrow\quad\quad\quad\quad\quad\\ &= 4x(x-3)(x+3)=0\\ \Rightarrow\quad\quad\quad\quad\quad\\ & x =0 \quad \text {or} \quad (x-3) =0 \quad \text {or} \quad (x+3) =0 \\ \Rightarrow\quad\quad\quad\quad\quad\\ & x =0 \quad \text {or} \quad x =3 \quad \text {or} \quad x=-3 \end{aligned} So, critical numbers are: $0$, $-3$ and $3$. Now, evaluate the function at the critical numbers and the endpoints: $$\text{when } x=-4: f(-4)=(-4)^{4}-18(-4)^{2}+1 =-31 \\ \text{when } x=-3: f(-3)=(-3)^{4}-18(-3)^{2}+1 =-80 \\ \text{when } x=0: f(0)=(0)^{4}-18(0)^{2}+1 =1 \\ \text{when } x=3: f(3)=(3)^{4}-18(3)^{2}+1 =-80 \\ \text{when } x=4: f(4)=(4)^{4}-18(4)^{2}+1 =-31 \\$$ We conclude that, the absolute maximum, $1$, occurs when $x=0$ and the absolute minimum, $-80$ occurs when $x=-3, 3.$