in exercises 17 - 26, find the extreme values...
in exercises 17 - 26, find the extreme values of the function on the given interval. 18. f(x)=x³ - 9/2x² - 30x + 3 on 0,6.
Answer
# Answer:
The maximum value is \(3\) at \(x = 0\), and the minimum value is \(-177\) at \(x = 5\).
# Explanation:
## Step1: Find the derivative
Differentiate \(f(x)=x^{3}-\frac{9}{2}x^{2}-30x + 3\) using power - rule. \(f'(x)=3x^{2}-9x - 30\).
## Step2: Set the derivative equal to zero
\(3x^{2}-9x - 30 = 0\). Divide through by \(3\): \(x^{2}-3x - 10=0\). Factor: \((x - 5)(x+2)=0\). Solutions are \(x = 5\) and \(x=-2\). But \(x=-2\) is outside the interval \([0,6]\), so we discard it.
## Step3: Evaluate the function at critical and end - points
Evaluate \(f(x)\) at \(x = 0\), \(x = 5\), and \(x = 6\).
\(f(0)=0^{3}-\frac{9}{2}(0)^{2}-30(0)+3 = 3\).
\(f(5)=5^{3}-\frac{9}{2}(5)^{2}-30(5)+3=125-\frac{225}{2}-150 + 3=125-112.5-150 + 3=-177\).
\(f(6)=6^{3}-\frac{9}{2}(6)^{2}-30(6)+3=216 - 162-180 + 3=-123\).