Answer to Question #272426 in Calculus for Kia

Question

Absolute extrema in closed interval

a. h(x)=x²-2x, [0,4]

b. f(x)= (2x+5)/3, [0,5]

Answer 1

a)

h(x)=x²-2x, [0,4]

Differentiating with respect to x

$\frac{dh}{dx} = 2x -2$

For maximum or minimum values of h(x) , $\frac{dh}{dx} =0$

=> 2x - 2 = 0

=> x = $\frac{2}{2} =1$

Differentiating again with respect to x

$\frac{d²h}{dx²} = 2 > 0$ for every value of x

So h(x) has a local minimum at x= 1

The absolute minimum value of h(x) is

Min {h(0), h(1),h(4)}

= Min {0, (1²-2.1), (4² - 2.4)}

= Min{0, -1 , 8}

= -1

Absolute maximum value of h(x) in [0,4] is

Max { h(0), h(1),h(4)}

= Max{0, (1²-2.1), (4²-2.4)}

= Max { 0, -1, 8}

= 8

b)

f(x)= $\frac{(2x+5)}{3}$ , [0,5]

=> f(x) = $\frac{2x}{3}+\frac{5}{3}$ , [0,5]

Differentiating with respect to x

$\frac{df}{dx} = \frac{2}{3}$ ≠ 0

$\frac{d²f}{dx²} =$ 0

So f(x) has neither local maximum nor local minimum value.

Absolute minimum of f(x) in [0,5] is

Min { f(0), f(5)}

= Min {${\frac{5}{3}, \frac{15}{3}}$ }

= Min{ ${\frac{5}{3}, 5}$ }

= $\frac{5}{3}$

Absolute maximum of f(x) in [0,5] is

Max{f(0), f(5)}

= Max{${\frac{5}{3}, \frac{15}{3}}$ }

=Min{${\frac{5}{3}, 5}$ }

= 5