Question

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

enter the minimum value here, or enter none if there isnt one.

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 \(x=-9
otin[0,10]\), 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 \(-1123\).