How do you find the relative extrema for #f(x) = x - log_4x#?

Question

Paisley Johnson · Accepted Answer

The minimum of #f(x)# is# f(1/(ln 4))=0.957#, nearly.

Use #log_b a = log_c a log_a c# #ln x=log_e x=log_4 x log_e 4=log_4 x ln 4#. #log_4 x=ln x/ln 4#. Now, #f(x)=x-ln x/ln 4# #f'=1-1/(x ln 4)=0#, when #x=1/ln 4# #f''=1/(x^2 ln 4)>0#, for all x, as #ln 4=1.3863>0#.. Thus,# f(1/ln 4)# is the minimum of f(x). #f(1/ln 4)=1/ln 4-ln(1/ln 4)/ln 4=(1+ln(ln 4))/ln 4=0.957#, nearly.-

Charles Anctil · Answer

undefined To find the relative extrema for $ f(x) = x - \log_4(x) $, follow these steps:

1. Take the derivative of $ f(x) $ with respect to $ x $.
2. Set the derivative equal to zero and solve for $ x $.
3. Check the second derivative at each critical point to determine whether it's a relative minimum, maximum, or neither.

Now, let's go through these steps:

1. The derivative of $ f(x) = x - \log_4(x) $ is:
   $$ f'(x) = 1 - \frac{1}{x \ln(4)} $$

2. Setting $ f'(x) $ equal to zero and solving for $ x $:
   $$ 1 - \frac{1}{x \ln(4)} = 0 $$
   $$ 1 = \frac{1}{x \ln(4)} $$
   $$ x \ln(4) = 1 $$
   $$ x = \frac{1}{\ln(4)} $$

3. To determine the nature of the critical point, we need to check the second derivative:
   $$ f''(x) = \frac{1}{x^2 \ln^2(4)} $$
   Substituting $ x = \frac{1}{\ln(4)} $ into $ f''(x) $, we get:
   $$ f''\left(\frac{1}{\ln(4)}\right) = \frac{1}{\left(\frac{1}{\ln(4)}\right)^2 \ln^2(4)} = \ln^2(4) $$

Since the second derivative is positive, $ x = \frac{1}{\ln(4)} $ corresponds to a relative minimum.

Therefore, the relative minimum for $ f(x) = x - \log_4(x) $ occurs at $ x = \frac{1}{\ln(4)} $.