What are the extrema of #f(x)=f(x)= x^2 -4x +3#?

1 Answer
Jan 21, 2016

Answer:

The extrema is at x = 2; obtained by solving #f'(x) = 0#
#f'(x) = 2x -4 = 0#; Take a look at the graph it will help.

graph{x^2-4x+3 [-5, 5, -5, 5]} solve for x.

Explanation:

You would typically find the first derivative and second derivative to find the extrema, but in this case it is trivial simply find the first derivative. WHY? you should be able to answer this
Given #f(x) = x^2 - 4x + 3; f'(x) = 2x -4; f'' = 2# constant
Now set #f'(x) = 0 # and solve for ==> #x = 2#