f(x) = (x+1)/((x^2+1)^(1/2)) Need to find max's, min's and when f(x) increases most rapidly.?

Question

I know this involves the first and second derivatives and their zeros, but I keep getting stuck. Need help.

Answer 1

To start, first take the derivative of a simpler function. You'll need this for the one in question.

g(x) = √(x² + 1)
g'(x) = (1/2)(2x)/√(x² + 1) = x/√(x² + 1)

With this in mind, let's proceed.

f(x) = (x+1)/√(x² + 1)

f'(x) = {(1)√(x² + 1) - [x/√(x² + 1)](x + 1)}/(x² + 1)

Multiply numerator and denominator by √(x² + 1)

f'(x) = {(x² + 1) - x(x + 1)}/{(x² + 1)^(3/2)}
f'(x) = {x² + 1 - x² - x}/{(x² + 1)^(3/2)}
f'(x) = (1 - x)/{(x² + 1)^(3/2)}

Set this equal to zero to find the critical value(s).

Answer 2

This is just derivatives and quadratic equations. Solve for x:

d[(x+1)*(x^2+1)^-.5]
------------------------- = 0;
dx

The derivative part is: "the derivative of the first term times the second + the derivative of the second term times the first," or, in mathematics terms:

(x^2+1)^-.5 + (x+1)*(-.5*(x^2+1)^-1.5*2x

simplified, this is:

1/(x^2-1)^.5 * [1-(x^2+x)/(x^2+1)]

Solve for x by setting this equal to 0.

Answer 3

69

Answer 4

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