find the extreme values of the function on the given interval.\n$f(x)=x^{3}+(\\frac{3}{2})x^{2}-216x - 3$ on…

find the extreme values of the function on the given interval.\n$f(x)=x^{3}+(\\frac{3}{2})x^{2}-216x - 3$ on $0,10$\nenter the maximum value here, or enter none if there isnt one.\n\nenter the minimum value here, or enter none if there isnt one.

find the extreme values of the function on the given interval.\n$f(x)=x^{3}+(\\frac{3}{2})x^{2}-216x - 3$ on $0,10$\nenter the maximum value here, or enter none if there isnt one.\n\nenter the minimum value here, or enter none if there isnt one.

Answer

Explanation:

Step1: Find the derivative of the function

Using the power rule ( (x^n)^\prime=nx^{n - 1}), for (f(x)=x^{3}+\frac{3}{2}x^{2}-216x - 3), the derivative (f^\prime(x)=3x^{2}+3x-216). Factor out 3: (f^\prime(x)=3(x^{2}+x - 72)). Factor the quadratic: (f^\prime(x)=3(x + 9)(x - 8)).

Step2: Find the critical points

Set (f^\prime(x)=0), then (3(x + 9)(x - 8)=0). Solutions are (x=-9) and (x = 8). Since (x=-9\notin[0,10]), we consider (x = 8).

Step3: Evaluate the function at the critical point and endpoints

  • Evaluate (f(x)) at (x = 0): (f(0)=0^{3}+\frac{3}{2}(0)^{2}-216(0)-3=-3).
  • Evaluate (f(x)) at (x = 8): (f(8)=8^{3}+\frac{3}{2}(8)^{2}-216(8)-3=512+\frac{3}{2}\times64-1728 - 3=512 + 96-1728-3=-1123).
  • Evaluate (f(x)) at (x = 10): (f(10)=10^{3}+\frac{3}{2}(10)^{2}-216(10)-3=1000+\frac{3}{2}\times100-2160 - 3=1000 + 150-2160-3=-1013).

Answer:

The maximum value is (-3). The minimum value is (-1123).