Question

find max value of function f(x)=1-x-x^2

Answer 1

frst differntiate

Answer 2

The max. value of a parabola is at its vertex.

Answer 3

f'=-1-2x=0
-1=2x
x=-1/2
now again differntiate f
f''=-2
f''<0 at x=-1/2
so it ismax at x=-1/2
find F(-1/2)

Answer 4

The x-coordinate of the vertex is -b/2a = -1/2.

Answer 5

so the formula is -b/2a?

Answer 6

Actually, the y at vertex is -D/4a where D is the discriminant.

Answer 7

im confused with the first response where did the -2x come from

Answer 8

Hi @Ssusu, to answer your question, the -2x came from taking the derivative (using Calculus). Don't Panic! You can find the maximum of this function without using Calculus. ;)

Answer 9

oh ok and how do I do that

Answer 10

To take the derivative requires a bit of prerequisite learning (pre-calculus). First, we would define a mathematical limit. Then we would take the limit of the difference quotient of that function.

Answer 11

It would probably be best to take a Calculus math class to really learn these concepts, but with the internet you could teach yourself or watch some Khan Academy videos.

Answer 12

http://www.youtube.com/watch?v=rAof9Ld5sOg

Answer 13

this question is for my pre cal hw chapter 2

Answer 14

If you are in pre-cal class, my best guess would be solving the problem by completing square, that is to rearrange the equation into the form y=a(x-h)^2+k.
The vertex is (h,k), with k being the maximum or minimum value and k be the corresponding x value.
f(x) =y =1-x-x^2
= - (x^2 + x) +1
= - (x + 1/2)^2 + (1/2)^2 + 1
= - ( x + 1/2 )^2 + 5/4
Now, you can find the vertex, and hence the maximum value.