find the critical points of the following function on the given interval. identify the absolute maximum and…

find the critical points of the following function on the given interval. identify the absolute maximum and absolute minimum values (if they exist)\n$f(x)=3x^{3}-36x^{2}$ on $-1,9$
Answer
Explanation:
Step1: Find the derivative of ( f(x) )
Given ( f(x)=3x^{3}-36x^{2} ), using the power rule ( (x^{n})^\prime = nx^{n - 1} ), we have ( f^\prime(x)=9x^{2}-72x=9x(x - 8) ).
Step2: Find the critical points
Set ( f^\prime(x)=0 ), so ( 9x(x - 8)=0 ). Solving this equation:
- When ( 9x=0 ), we get ( x = 0 ).
- When ( x - 8=0 ), we get ( x = 8 ). Both ( x = 0 ) and ( x = 8 ) are in the interval ( [-1,9] ).
Step3: Evaluate the function at critical points and endpoints
- Evaluate ( f(x) ) at ( x=-1 ): ( f(-1)=3\times(-1)^{3}-36\times(-1)^{2}=-3 - 36=-39 ).
- Evaluate ( f(x) ) at ( x = 0 ): ( f(0)=3\times0^{3}-36\times0^{2}=0 ).
- Evaluate ( f(x) ) at ( x = 8 ): ( f(8)=3\times8^{3}-36\times8^{2}=3\times512-36\times64=1536 - 2304=-768 ).
- Evaluate ( f(x) ) at ( x = 9 ): ( f(9)=3\times9^{3}-36\times9^{2}=3\times729-36\times81=2187 - 2916=-729 ).
Answer:
- Critical points: ( x = 0,8 )
- Absolute minimum: occurs at ( x = 8 )
- Absolute maximum: occurs at ( x = 0 )