Question

find the critical points of the following function on the given interval. identify the absolute maximum and absolute minimum values (if they exist)
$f(x)=3x^{3}-36x^{2}$ on $-1,9$

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 \( x = 0 \)