OpenStudy (anonymous):

quick question find max val and min value of on [0,5]

3 years ago
OpenStudy (anonymous):

i differentiated and got -2x-4=0

3 years ago
OpenStudy (anonymous):

found a critical point at x=-2

3 years ago
OpenStudy (anonymous):

i guess you have a parabola, so perhaps $$2$$ is the first coordinate of the vertex

3 years ago
OpenStudy (anonymous):

since you parabola evidently opens down, that is a maximum

3 years ago
OpenStudy (anonymous):

wait no the function is 12-4x-x^2

3 years ago
OpenStudy (anonymous):

the stuff i wrote is the f'(x) of it

3 years ago
OpenStudy (anonymous):

and solved for x

3 years ago
OpenStudy (anonymous):

so i should evaluate at the end points and the critical point ,but -2 is not in the domain so i shouldnt use it right?

3 years ago
OpenStudy (anonymous):

yeah, you don't need calculus to find the vertex of a quadratic first coordinate is $$-\frac{b}{2a}$$ which in your case is $$-2$$

3 years ago
OpenStudy (anonymous):

you made a mistake solving $-4-2x=0$ you should get $-4=2x\\ -2=x$

3 years ago
OpenStudy (anonymous):

oh wait, you did get $$-2$$ i read it incorrectly

3 years ago
OpenStudy (anonymous):

yeah thats what i got

3 years ago
OpenStudy (anonymous):

but my question is i shouldnt use that value because is not part of the domain right?

3 years ago
OpenStudy (anonymous):

ok so as you said, since $$-2$$ is not in the domain, just check the endpoints 0 will give a max, 5 a min

3 years ago
OpenStudy (anonymous):

right, don't use it, it is not in the domain

3 years ago
OpenStudy (anonymous):

parabola is decreasing on $$(-2,\infty)$$ so it will be largest at $$0$$ and smallest at $$5$$

3 years ago
OpenStudy (anonymous):

cool thanks

3 years ago
OpenStudy (anonymous):

ye

3 years ago
OpenStudy (anonymous):

oops i means yw

3 years ago