What is the extreme f(x)=(sqrt(2-x^2))/(2x+1)+7 ?

The extreme f(x)=(sqrt(2-x^2))/(2x+1)+7 is Maximum(-sqrt(2),7),Minimum(sqrt(2),7)

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extreme f(x)=(sqrt(2-x^2))/(2x+1)+7

extreme

Maximum (- 2, 7), Minimum (2, 7)

Solution steps

f^{'} (x) = \frac{- x - 4}{( 2 x + 1 ) ^{2} - ( x + 2 ) ( x - 2 )}

Find intervals:

Decreasing

: - 2 < x < - \frac{1}{2},

Decreasing

: - \frac{1}{2} < x < 2

Plug

x = - 2

into

\frac{2 - x ^{2}}{2 x + 1} + 7 : 7

Maximum (- 2, 7)

Plug

x = 2

into

\frac{2 - x ^{2}}{2 x + 1} + 7 : 7

Minimum (2, 7)

Maximum (- 2, 7), Minimum (2, 7)

What is the extreme f(x)=(sqrt(2-x^2))/(2x+1)+7 ?
The extreme f(x)=(sqrt(2-x^2))/(2x+1)+7 is Maximum(-sqrt(2),7),Minimum(sqrt(2),7)