section 2.8: optimization (homework)\nscore: 49.5/120 answered: 4/12\nquestion 6\nconsider the function…

section 2.8: optimization (homework)\nscore: 49.5/120 answered: 4/12\nquestion 6\nconsider the function (f(x)=1 - 7x^{2}), (-4leq xleq1).\nthe absolute maximum value is \nand this occurs at (x=)\nthe absolute minimum value is \nand this occurs at (x=)\nquestion help: video\nsubmit question

section 2.8: optimization (homework)\nscore: 49.5/120 answered: 4/12\nquestion 6\nconsider the function (f(x)=1 - 7x^{2}), (-4leq xleq1).\nthe absolute maximum value is \nand this occurs at (x=)\nthe absolute minimum value is \nand this occurs at (x=)\nquestion help: video\nsubmit question

Answer

Explanation:

Step1: Find the derivative

Differentiate $f(x)=1 - 7x^{2}$ using power - rule. The derivative $f'(x)=-14x$.

Step2: Find the critical points

Set $f'(x) = 0$. So, $-14x=0$, which gives $x = 0$.

Step3: Evaluate the function at critical point and endpoints

Evaluate $f(x)$ at $x=-4,x = 0,x = 1$. For $x=-4$, $f(-4)=1-7\times(-4)^{2}=1 - 112=-111$. For $x = 0$, $f(0)=1-7\times0^{2}=1$. For $x = 1$, $f(1)=1-7\times1^{2}=1 - 7=-6$.

Answer:

The absolute maximum value is $1$ and this occurs at $x = 0$. The absolute minimum value is $-111$ and this occurs at $x=-4$.