Question

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

Answer 1

i differentiated and got -2x-4=0

Answer 2

found a critical point at x=-2

Answer 3

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

Answer 4

since you parabola evidently opens down, that is a maximum

Answer 5

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

Answer 6

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

Answer 7

and solved for x

Answer 8

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?

Answer 9

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

Answer 10

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

Answer 11

oh wait, you did get \(-2\) i read it incorrectly

Answer 12

yeah thats what i got

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

cool thanks

Answer 18

ye

Answer 19

oops i means yw