What is the extreme f(x)=sqrt(-x^2+6x+16) ?

The extreme f(x)=sqrt(-x^2+6x+16) is Minimum(-2,0),Maximum(3,5),Minimum(8,0)

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extreme f(x)=sqrt(-x^2+6x+16)

extreme points

f (x) = - x^{2} + 6 x + 16

Minimum (- 2, 0), Maximum (3, 5), Minimum (8, 0)

Solution steps

f^{'} (x) = \frac{- x + 3}{- x ^{2} + 6 x + 16}

Find intervals:

Increasing

: - 2 < x < 3,

Decreasing

: 3 < x < 8

Plug

x = - 2

into

- x^{2} + 6 x + 16 : 0

Minimum (- 2, 0)

Plug

x = 3

into

- x^{2} + 6 x + 16 : 5

Maximum (3, 5)

Plug

x = 8

into

- x^{2} + 6 x + 16 : 0

Minimum (8, 0)

Minimum (- 2, 0), Maximum (3, 5), Minimum (8, 0)

f (x) = 4^{x - 2}

3 x 4 x^{2} + 2

f (x) = x + 3 - 2

f (x) = \frac{x - 5}{x + 1}

f (x) = \frac{4}{3 x + 1}

What is the extreme f(x)=sqrt(-x^2+6x+16) ?
The extreme f(x)=sqrt(-x^2+6x+16) is Minimum(-2,0),Maximum(3,5),Minimum(8,0)