How do you find the max or minimum of #f(x)=x^2-x-6#?

1 Answer
Dec 6, 2016

The minimum of #f(x)# is at #(1/2,-25/4)#

Explanation:

Let 's find the derivative of #f(x)#

#f(x)=x^2-x-6#

#f'(x)=2x-1#

#f'(x)=0# when #x=1/2#

Let's do a chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##1/2##color(white)(aaaa)##+oo#

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

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##darr##color(white)(aaaa)##uarr#

So it's a minimum.

We can also calculate the second derivative

#f''(x)=2>0#, so it'a a minimum

graph{x^2-x-6 [-14.24, 14.23, -7.12, 7.12]}