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

Question

Evelyn Agard · Accepted Answer

#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)#

Axel Alvarez · Answer

undefined To find the critical points of $ f(x) = \log_7(1+x^2) $, we need to find where its derivative is equal to zero or undefined.

First, let's find the derivative of $ f(x) $:
$$ f'(x) = \frac{d}{dx} \left( \log_7(1+x^2) \right) $$

Using the chain rule and the derivative of the natural logarithm, we have:
$$ f'(x) = \frac{1}{\ln(7)} \cdot \frac{1}{1+x^2} \cdot \frac{d}{dx}(1+x^2) $$

$$ f'(x) = \frac{2x}{(1+x^2) \ln(7)} $$

Now, to find the critical points, set the derivative equal to zero and solve for $ x $:
$$ \frac{2x}{(1+x^2) \ln(7)} = 0 $$

$$ 2x = 0 $$

$$ x = 0 $$

So, the only critical point of $ f(x) = \log_7(1+x^2) $ is $ x = 0 $.