Answer:
The absolute minimum value is "[tex]-\frac{21}{4}[/tex]" and the absolute maximum value is "[tex]15[/tex]".
Step-by-step explanation:
Given:
[tex]f(x)=x^2-x-5[/tex]
on,
[tex][0,5][/tex]
By differentiating it, we get
⇒ [tex]f'(x)=2x-1[/tex]
Set [tex]f'(x)=0[/tex]
then,
⇒ [tex]2x-1=0[/tex]
[tex]2x=1[/tex]
[tex]x=\frac{1}{2}[/tex] (Critical point)
When x=0,
⇒ [tex]f(x)=-5[/tex]
When [tex]x=\frac{1}{2}[/tex],
⇒ [tex]f(x)=-\frac{21}{4}[/tex] (Absolute minimum)
When [tex]x=5[/tex]
⇒ [tex]f(x)=15[/tex] (Absolute maximum)
