Question

consider the function $f(x) = 5x^2 - 8x + 2$, $0 \leq x \leq 6$. the absolute maximum of $f(x)$ (on the given interval) is at $x = $ and the absolute minimum of $f(x)$ (on the given interval) is at $x = $

Explanation:

Step1: Find critical points

The function is \( f(x) = 5x^2 - 8x + 2 \). The derivative \( f'(x)=10x - 8 \). Set \( f'(x) = 0 \), so \( 10x - 8 = 0 \), solving gives \( x=\frac{8}{10}=\frac{4}{5}=0.8 \).

Step2: Evaluate function at critical point and endpoints

The interval is \( 0\leq x\leq6 \). Evaluate \( f(x) \) at \( x = 0 \), \( x=\frac{4}{5} \), and \( x = 6 \).

• At \( x = 0 \): \( f(0)=5(0)^2 - 8(0)+2 = 2 \).
• At \( x=\frac{4}{5} \): \( f(\frac{4}{5})=5(\frac{4}{5})^2 - 8(\frac{4}{5})+2 = 5(\frac{16}{25})-\frac{32}{5}+2=\frac{16}{5}-\frac{32}{5}+2=-\frac{16}{5}+2=-\frac{6}{5}=-1.2 \).
• At \( x = 6 \): \( f(6)=5(6)^2 - 8(6)+2 = 5(36)-48 + 2=180 - 48+2 = 134 \).

Step3: Compare values

The largest value is \( 134 \) at \( x = 6 \), and the smallest value is \( -1.2 \) at \( x=\frac{4}{5} \).

Answer:

The absolute maximum of \( f(x) \) (on the given interval) is at \( x = \boldsymbol{6} \) and the absolute minimum of \( f(x) \) (on the given interval) is at \( x = \boldsymbol{\frac{4}{5}} \) (or \( 0.8 \)).