Find the absolute maximum and absolute minimum values of f on the given interval.

Question

x/(x^2 - x +16), [0,12]?

Answer 1

(c)

The function is f(x) = x/(x²-x+16) and the interval is [0, 12].

Differentiate with respect to x.

f'(x) = [(x²-x+16)(d/dx)x-x(d/dx)(x²-x+16)] / (x²-x+16)²

f'(x) = [(x²-x+16)-x(2x-1)] / (x²-x+16)²

f'(x) = [(x²-x+16)-2x²+x)] / (x²-x+16)²

f'(x) = (-x²+16) / (x²-x+16)²

To find the critical numbers.

Set f'(x) = 0 then find x values.and Find at which values of x  f'(x) does not exist.

f'(x) = 0 ⇒ (-x²+16) / (x²-x+16)² =0

f'(x) = 0 ⇒ 16 - x² = 0 ⇒ x = ±4.

x = -4 does not lie in the interval [0 , 12], so this is negligible.

x = 4 is lie in the interval [0 , 12], so we can consider it.

To find at which values of x  f'(x) does not exist when (x²-x+16)² = 0

x²-x+16 = 0

x =  { -(-1) ± √[(-1)²-4×1×16] } / 2×1

x =  {1 ± √[-63] } / 2

x = (1 ± i 7.937 ) / 2

x =  0.5 ± i 3.97

The imaginary numbers are negligible.

Left Endpoint: f(0) = 0/(0²-0+16)) = 0 [ Minimum ].

Critical Number: f(4) = 4/(4²-4+16)) = 4/28 = 0.143 [ Maximum ].

Right Endpoint: f(12) = 12/(12²-12+16)) = 12/148 = 0.081

Solution :

Absolute Maximum = 0.143

Absolute Minimum = 0