Question

consider the function ( f(x) = 2x^{3}+9x^{2}-168x + 1), ( -7leq xleq5).
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 derivative of \(f(x)=2x^{3}+9x^{2}-168x + 1\) is \(f^\prime(x)=6x^{2}+18x-168\).
Factor \(f^\prime(x)\): \(f^\prime(x)=6(x^{2}+3x - 28)=6(x + 7)(x-4)\).

Step2: Find the critical points

Set \(f^\prime(x)=0\), then \(6(x + 7)(x - 4)=0\).
Solving \(x+7 = 0\) gives \(x=-7\), and solving \(x - 4=0\) gives \(x = 4\). Both \(x=-7\) and \(x = 4\) are in the interval \([-7,5]\).

Step3: Evaluate the function at the critical points and endpoints

• Evaluate \(f(-7)\):

\(f(-7)=2(-7)^{3}+9(-7)^{2}-168(-7)+1=2(-343)+9(49)+1176 + 1=-686+441+1176+1=932\).

• Evaluate \(f(4)\):

\(f(4)=2(4)^{3}+9(4)^{2}-168(4)+1=2(64)+9(16)-672 + 1=128+144-672+1=-399\).

• Evaluate \(f(5)\):

\(f(5)=2(5)^{3}+9(5)^{2}-168(5)+1=2(125)+9(25)-840+1=250+225-840 + 1=-364\).

Answer:

The absolute minimum value is \(-399\) and the absolute maximum value is \(932\).