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

The extreme f(x)=(x+6)/(sqrt(x^2+6)) is Maximum(1,sqrt(7))

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extreme f(x)=(x+6)/(sqrt(x^2+6))

extreme

Maximum (1, 7)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = 1

into

\frac{x + 6}{x ^{2} + 6} : 7

Maximum (1, 7)

What is the extreme f(x)=(x+6)/(sqrt(x^2+6)) ?
The extreme f(x)=(x+6)/(sqrt(x^2+6)) is Maximum(1,sqrt(7))