How do you find the local max and min for $f(x) = 1 - sqrt(x)$ ?

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Answer 1 · 2016-12-15T03:43:48

Local maximum is $1$ (at $x=0$ ).

$f'(x) = -1/(2sqrtx)$

$f'(x)$ is never $0$ and is undefined at $x=0$ which is int he domain of $f$ . So $0$ is a critical number.

$f'$ is not defined left of $0$ and is negative $right of$ 0 $, so$ f(0) = 1# is a local maximum.

There are no other critical numbers, so there are no other local extrema.

How do you find the local max and min for f(x) = 1 - sqrt(x)f(x) = 1 - sqrt(x)?