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

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

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

extreme points

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) = x^{3} - 6 x^{2} + 8 x

f (x) = x^{3} - 12 x + 1

\frac{4 x + 9}{3 x - 2}

x + 2 y = 16

x^{2} - 2 x - 35

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