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

Textbook Question

Chapter 3, 3.4 - Problem 38 - Calculus of a Single Variable (10th Edition, Ron Larson).
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thilina-g's profile pic

thilina-g | College Teacher | (Level 1) Educator

Posted on

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.

loves2learn's profile pic

loves2learn | (Level 3) Salutatorian

Posted on

Find the first derivative.

Given,

`f(x)=sqrt(x^2+1) `

So,

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

This simplifies,

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

Now we need to find the value(s) of x that make the first derivative zero to find the critical points.

`0=x ` (critical point)

Note that the denominator can never equal 0

This will be a relative extrema if it changes sign, so find 2 values around it to test using the first derivative test, like -1 and 1

`f'(-1)=-1/sqrt2 `

`f'(1)=1/sqrt2 `

Therefore, this a relative extrema. The entire point is `(0,1)`

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