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

Nov 16, 2015

$P \left(0 , 0\right)$

#### Explanation:

We have that

${\log}_{7} \left(x\right) = \ln \frac{x}{\ln} \left(7\right)$

So we have

$y = {\log}_{7} \left(1 + {x}^{2}\right) = \ln \frac{1 + {x}^{2}}{\ln} \left(7\right)$

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

$y = \ln \frac{u}{\ln} \left(7\right)$

And so,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\ln} \left(7\right) \cdot \frac{d}{\mathrm{du}} \ln \left(u\right) \cdot \frac{d}{\mathrm{dx}} \left(1 + {x}^{2}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\ln} \left(7\right) \cdot \frac{1}{u} \cdot 2 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x}{\ln \left(7\right) \cdot \left(1 + {x}^{2}\right)}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0 = \frac{2 x}{\ln \left(7\right) \cdot \left(1 + {x}^{2}\right)}$

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

$0 = 2 x$
$x = 0$

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