SOLUTION: f(x)=2-2x^2 Estimate the intervals on which the function is increasing or decreasing and estimate any relative maxima or minima. Thanks and could you also show work.

Algebra ->  Graphs -> SOLUTION: f(x)=2-2x^2 Estimate the intervals on which the function is increasing or decreasing and estimate any relative maxima or minima. Thanks and could you also show work.       Log On


   

Question 608072: f(x)=2-2x^2
Estimate the intervals on which the function is increasing or decreasing and estimate any relative maxima or minima. Thanks and could you also show work.
Found 2 solutions by ewatrrr, nerdybill:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi

f(x)=2-2x^2 

f'(x) = -4x 

f'(x)>0  Increasing (−∞,0)

f'(x)<0  Decreasing  (0,∞)

The relative maxima for f(x)= highlight%282%29(when f'=0, x = 0)

Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
f(x)=2-2x^2
Estimate the intervals on which the function is increasing or decreasing and estimate any relative maxima or minima. Thanks and could you also show work
.
This is a "quadratic" because it is a polynomial of degree 2.
Because it's a quadratic, it is a parabola.
We know the parabola opens downward because the coefficient associated with the x^2 term is negative (think sad face).
Since it is a parabola that opens downwards, we know the vertex is at the MAXIMUM.
.
Axis of symmetry:
x = -b/(2a)
x = -0/(2(-2))
x = -0/(-4)
x = 0
.
Increasing interval:
(-oo, 0)
Decreasing interval:
(0, +oo)
where oo is for infinity
.
Maximum:
f(0)=2-2(0)^2
f(0)=2
so, max is at (0,2)