Question

section 2.8: optimization (homework)
score: 69.5/120 answered: 6/12
question 8
consider the function (f(x)=x^{4}-50x^{2}+1), (-4leq xleq11).
this function has an absolute minimum value equal to
and an absolute maximum value equal to
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Explanation:

Step1: Find the derivative of the function

Differentiate $f(x)=x^{4}-50x^{2}+1$ using the power - rule. $f^\prime(x)=4x^{3}-100x = 4x(x^{2}-25)=4x(x - 5)(x + 5)$.

Step2: Find the critical points

Set $f^\prime(x)=0$. Then $4x(x - 5)(x + 5)=0$. The critical points are $x=-5,x = 0,x = 5$. But $x=-5$ is outside the interval $[-4,11]$, so we discard it. We are left with $x = 0$ and $x = 5$.

Step3: Evaluate the function at the critical points and endpoints

Evaluate $f(x)$ at $x=-4,x = 0,x = 5,x = 11$.
$f(-4)=(-4)^{4}-50(-4)^{2}+1=256-800 + 1=-543$.
$f(0)=0^{4}-50\times0^{2}+1=1$.
$f(5)=5^{4}-50\times5^{2}+1=625-1250 + 1=-624$.
$f(11)=11^{4}-50\times11^{2}+1=14641-6050+1=8592$.

Answer:

The absolute minimum value is $-624$ and the absolute maximum value is $8592$.