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
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