Question

find the absolute maximum and absolute minimum values of f on the given interval.

f(x)=6x^{4}-8x^{3}-24x^{2}+1, -2,3

absolute minimum value

absolute maximum value

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Explanation:

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

Using the power rule \( (x^n)^\prime=nx^{n - 1} \), we have \( f^\prime(x)=24x^{3}-24x^{2}-48x=24x(x^{2}-x - 2)=24x(x - 2)(x+1) \)

Step2: Find the critical points

Set \( f^\prime(x)=0 \), then \( 24x(x - 2)(x + 1)=0 \). Solving for \( x \), we get \( x=-1,x = 0,x = 2 \). All of these critical points \( x=-1,x = 0,x = 2 \) are in the interval \( [-2,3] \)

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

• For \( x=-2 \): \( f(-2)=6\times(-2)^{4}-8\times(-2)^{3}-24\times(-2)^{2}+1=6\times16+8\times8-24\times4 + 1=96 + 64-96+1=65 \)
• For \( x=-1 \): \( f(-1)=6\times(-1)^{4}-8\times(-1)^{3}-24\times(-1)^{2}+1=6 + 8-24 + 1=-9 \)
• For \( x = 0 \): \( f(0)=6\times0^{4}-8\times0^{3}-24\times0^{2}+1=1 \)
• For \( x = 2 \): \( f(2)=6\times2^{4}-8\times2^{3}-24\times2^{2}+1=6\times16-8\times8-24\times4+1=96-64 - 96+1=-63 \)
• For \( x = 3 \): \( f(3)=6\times3^{4}-8\times3^{3}-24\times3^{2}+1=6\times81-8\times27-24\times9+1=486-216-216 + 1=55 \)

Answer:

The absolute minimum value is \(-63\) and the absolute maximum value is \(65\)