Question

consider the function $f(x)=x^{4}-72x^{2}+11$, $-5leq xleq13$. this function has an absolute minimum value equal to and an absolute maximum value equal to

Explanation:

Step1: Find the derivative of the function

The function is \(f(x)=x^{4}-72x^{2}+11\). Using the power rule \((x^{n})^\prime = nx^{n - 1}\), the derivative \(f^\prime(x)=4x^{3}-144x=4x(x^{2}-36)=4x(x - 6)(x + 6)\).

Step2: Find the critical points

Set \(f^\prime(x)=0\). Then \(4x(x - 6)(x + 6)=0\). Solving for \(x\), we get \(x = 0\), \(x=6\), \(x=-6\). But since \(-5\leq x\leq13\), we discard \(x=-6\). So the critical points in the interval \([-5,13]\) are \(x = 0\) and \(x = 6\).

Step3: Evaluate the function at the critical points and endpoints

• When \(x=-5\): \(f(-5)=(-5)^{4}-72(-5)^{2}+11=625-1800 + 11=-1164\).
• When \(x = 0\): \(f(0)=0^{4}-72\times0^{2}+11 = 11\).
• When \(x = 6\): \(f(6)=6^{4}-72\times6^{2}+11=1296-2592+11=-1285\).
• When \(x = 13\): \(f(13)=13^{4}-72\times13^{2}+11=28561-12168+11=16404\).

Answer:

The absolute minimum value is \(-1285\) and the absolute maximum value is \(16404\).