What are the extrema of #f(x)=(x - 4)(x - 5)# on #[4,5]#?

1 Answer
Nov 5, 2015

Answer:

The extremum of the function is (4.5 , -0.25)

Explanation:

#f(x) = (x-4)(x-5)# can be rewritten to #f(x) = x^2 - 5x - 4x + 20 = x^2-9x+20#.

If you derivate the function, you will end up with this:
#f'(x) = 2x - 9#.
If you don't how to derivate functions like these, check the description further down.

You want to know where #f'(x) = 0# , because that's where the gradient = 0.

Put #f'(x) = 0# ;
#2x - 9 = 0#
#2x = 9#
#x = 4.5#

Then put this value of x into the original function.
#f(4.5) = (4.5 - 4)(4.5-5)#
#f(4.5) = 0.5 * (-0.5)#
#f(4.5) = -0.25#

Crach course on how to derivate these types of functions:
Multiply the exponent with the base number, and decrease the exponent by 1.

Example:
#f(x) = 3x^3 - 2x^2 - 2x + 3#
#f'(x) = 3 * 3x^(3-1) - 2 * 2x^(2-1) - 1 * 2x^(1-1)#
#f'(x) = 9x^2 - 2x - 2x^0#
#f'(x) = 9x^2 - 2x - 2#