What is the extreme f(x)=sqrt(1-(x-3)^2) ?

The extreme f(x)=sqrt(1-(x-3)^2) is Minimum(2,0),Maximum(3,1),Minimum(4,0)

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extreme f(x)=sqrt(1-(x-3)^2)

extreme points

f (x) = 1 - (x - 3)^{2}

Minimum (2, 0), Maximum (3, 1), Minimum (4, 0)

Solution steps

f^{'} (x) = \frac{- x + 3}{- x ^{2} + 6 x - 8}

Find intervals:

Increasing

: 2 < x < 3,

Decreasing

: 3 < x < 4

Plug

x = 2

into

1 - (x - 3)^{2} : 0

Minimum (2, 0)

Plug

x = 3

into

1 - (x - 3)^{2} : 1

Maximum (3, 1)

Plug

x = 4

into

1 - (x - 3)^{2} : 0

Minimum (4, 0)

Minimum (2, 0), Maximum (3, 1), Minimum (4, 0)

f (x) = \frac{100}{x}

6 x - x^{2} + 7

y = \frac{3}{5}

f (x) = \frac{x}{x ^{2} + 16}

f (x) = lo g_{6} (4 x) + lo g_{6} (3)

What is the extreme f(x)=sqrt(1-(x-3)^2) ?
The extreme f(x)=sqrt(1-(x-3)^2) is Minimum(2,0),Maximum(3,1),Minimum(4,0)