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

1 Answer
Nov 16, 2015

Answer:

#P(0,0)#

Explanation:

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)#