# Locate all critical points (both types) of `h(x)= sqrt(1-7x^2)` The critical points are:

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Other critical points are the x-values that make the `(1/(2*sqrt (1-7x)^2)` undefined. Solving for x, we will get `x = +-sqrt (1/7)`. So, `+-sqrt (1/7)` are additional critical points.

`h(x)=sqrt(1-7x^2)` (i)

Differentiate (i) ,with respect to x

`h'(x)=(1/(2sqrt(1-7x^2)))(-14x)` (ii)

`h'(x)=0 if x=0`

Differentate (ii) with respect to x

`h''(x)={-7(sqrt(1-7x^2)-xd/(dx)(sqrt(1-7x^2))) } /(1-7x^2)`

``

`h''(x)}_{x=0}=-7<0`

`` Thus x=0 is a critical point and it is point of maxima.