How do you find the maximum and minimum values of the function #f(x)= x - ((64x)/(x+4))# on the interval [0,13]?

1 Answer
Jul 13, 2018

There is only a local minimum at #(12, -36)#

Explanation:

Calculate the first derivative of #f(x)#

The derivative of

#(x)'=1#

And by the quotient rule

#((64x)/(x+4))'=(64(x+4)-1(64x))/(x+4)^2#

#=256/(x+4)^2#

Therefore,

#f'(x)=1-256/(x+4)^2=(((x+4)^2)-64)/(x+4)^2#

#=(x^2+8x+16-256)/(x+4)^2#

#=(x^2+8x-240)/(x+4)^2#

#=((x+20)(x-12))/(x+4)^2#

The critical points are when

#f'(x)=0#

#=>#, #{(x=-20),(x=12):}#

Build a variation chart in the interval #[0,13]#

#color(white)(aaaa)##"Interval"##color(white)(aaaa)##(0,12)##color(white)(aaaa)##(12,13)#

#color(white)(aaaa)##"sign f'(x)"##color(white)(aaaaa)##-##color(white)(aaaaaaa)##+#

#color(white)(aaaa)##" f(x)"##color(white)(aaaaaaaa)##↘##color(white)(aaaaaaa)##↗#

There is only a local minimum at #(12, -36)#

graph{x-((64x)/(x+4)) [-73.8, 113.7, -63, 30.8]}