Question

consider the function ( f(x) = 5x^{2}-6x + 3, 0leq xleq9 ).
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 the derivative of the function

The derivative of \(f(x)=5x^{2}-6x + 3\) is \(f^\prime(x)=10x-6\).

Step2: Find the critical points

Set \(f^\prime(x) = 0\), so \(10x-6=0\), which gives \(x=\frac{6}{10}=\frac{3}{5}=0.6\).

Step3: Evaluate the function at critical and endpoint values

• At \(x = 0\): \(f(0)=5(0)^{2}-6(0)+3=3\).
• At \(x=\frac{3}{5}\): \(f(\frac{3}{5})=5(\frac{3}{5})^{2}-6(\frac{3}{5})+3=5\times\frac{9}{25}-\frac{18}{5}+3=\frac{9}{5}-\frac{18}{5}+3=\frac{9 - 18+15}{5}=\frac{6}{5}=1.2\).
• At \(x = 9\): \(f(9)=5(9)^{2}-6(9)+3=5\times81-54 + 3=405-54+3=354\).

Answer:

The absolute maximum of \(f(x)\) (on the given interval) is at \(x = 9\) and the absolute minimum of \(f(x)\) (on the given interval) is at \(x=\frac{3}{5}\).