# `f(x) = sqrt(x^2 + 1)` Find all relative extrema, use the second derivative test where applicable. First derivative,

`f'(x) = 1/2 * 2*x (x^2+1)^(1/2-1)`

`f'(x) = x(x^2+1)^(-1/2)`

Therefore, `x(x^2+1)^(-1/2) = 0.` This gives, x = 0

At x = 0, f(x) = 1.

Therefore, the point (0,1) is a critical point.

Second derivative test,

`f''(x) = 1*(x^2+1)^(-1/2) + x*(-1/2)*2x*(x^2+1)^(-1/2-1)`

`f''(x) = (x^2+1)^(-1/2) -x^2(x^2+1)^(-3/2)`

At x =...

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First derivative,

`f'(x) = 1/2 * 2*x (x^2+1)^(1/2-1)`

`f'(x) = x(x^2+1)^(-1/2)`

Therefore, `x(x^2+1)^(-1/2) = 0.` This gives, x = 0

At x = 0, f(x) = 1.

Therefore, the point (0,1) is a critical point.

Second derivative test,

`f''(x) = 1*(x^2+1)^(-1/2) + x*(-1/2)*2x*(x^2+1)^(-1/2-1)`

`f''(x) = (x^2+1)^(-1/2) -x^2(x^2+1)^(-3/2)`

At x = 0.

`f''(X) = (1)^(-1/2) - 0= 1 gt 0`

Therefore, the point (0,1) is a minimum.

Approved by eNotes Editorial Team