What are the critical points of $f (x) = log_7(1+x^2)$ ?

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Answer 1 · 2015-11-16T21:56:26

$P(0,0)$

We have that

$log_7(x) = ln(x)/ln(7)$

So we have

$y = log_7(1+x^2) = ln(1+x^2)/ln(7)$

Calling $1 + x^2 = u$ we have

$y = ln(u)/ln(7)$

And so,

$dy/dx = 1/ln(7)*d/(du)ln(u)*d/dx(1+x^2)$

$dy/dx = 1/ln(7)*1/u*2x$

$dy/dx = (2x)/(ln(7)*(1+x^2))$

A the critical points are those where the derivative is 0, or

$dy/dx = 0 = (2x)/(ln(7)*(1+x^2))$

The denominator will never become zero, so we'll never have any points the function isn't differentiable (in the positive reals), so

$0 = 2x$
$x = 0$

The function has 1 critical point, that is $P(0,0)$

What are the critical points of f (x) = log_7(1+x^2)f (x) = log_7(1+x^2)?