How do you find local maximum value of f using the first and second derivative tests: #f(x)= 5x - 3#?

Question

How do you find local maximum value of f using the first and second derivative tests:  #f(x)= 5x - 3#?

Christopher Agresta · Accepted Answer

I think you haven't any max or min here.

You can immediately recognize and say that your Linear function won't have a maximum. The graph of your function will be a straight line passing by #y=-3# and with slope #m=5#. Anyway, we can test our assumption by deriving once: #f'(x)=5# Now you can set your derivative equal to zero to find your max or min, but: #f'(x)=0# never happens because #f'(x)=5# that is never equal to zero! We do not have any max. If you derive again you would try to find the direction of the concavity (belly up or down) setting the second derivative greater than zero but this derivative will give you: #f''(x) =0# (derivative of #f'(x)=5#) that tells you that your function has no concavity (neither up nor down). Graphically: graph{5x-3 [-9.42, 10.58, 3.76, 13.76]}

Ezra Adams · Answer

undefined To find the local maximum value of $ f(x) = 5x - 3 $ using the first and second derivative tests:

1. Find the first derivative of $ f(x) $ with respect to $ x $.
   $ f'(x) = 5 $

2. Find the critical points by setting $ f'(x) $ equal to zero.
   $ 5 = 0 $ has no solutions, so there are no critical points.

3. Since there are no critical points, there are no local maximum values for $ f(x) $ within its domain.

Therefore, there is no local maximum value for $ f(x) = 5x - 3 $ using the first and second derivative tests.