Question

given the function ( f(x)=x^{4}-4x^{3}-3 ), determine the absolute maximum value of ( f ) on the closed interval ( -2,4 ).

Explanation:

Step1: Find the derivative of \(f(x)\)

Using the power rule \((x^n)^\prime=nx^{n - 1}\), for \(f(x)=x^{4}-4x^{3}-3\), the derivative \(f^\prime(x)=4x^{3}-12x^{2}=4x^{2}(x - 3)\)

Step2: Find the critical points

Set \(f^\prime(x)=0\), so \(4x^{2}(x - 3)=0\).
Solving \(4x^{2}(x - 3)=0\) gives \(x = 0\) or \(x=3\) (both \(x = 0\) and \(x = 3\) are in the interval \([-2,4]\))

Step3: Evaluate \(f(x)\) at the critical points and endpoints

• For \(x=-2\): \(f(-2)=(-2)^{4}-4(-2)^{3}-3=16 + 32-3=45\)
• For \(x = 0\): \(f(0)=0^{4}-4\times0^{3}-3=-3\)
• For \(x = 3\): \(f(3)=3^{4}-4\times3^{3}-3=81-108 - 3=-30\)
• For \(x = 4\): \(f(4)=4^{4}-4\times4^{3}-3=256-256-3=-3\)

Answer:

The absolute maximum value of \(f(x)\) on the interval \([-2,4]\) is \(45\)