find the absolute maximum and absolute minimum values of f on the given interval.\n\nf(x)=6x^{4}-8x^{3}-24x^{…

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)