Question

if f(X)= 2X-X^2,if x<1
2x^2-4x+3, if >or equal to 1
find two values of c in the interval [0,2] that satisfy the mean value theorem.


Please help I have no idea how to solve

Answer 1

when x = 2, y = 2(4)-8+3 = +3
when x = 0, y = 2(0) -0^2 = 0
delta y/delta x = (3-0)/2 = 1.5

so where in there does f' = 1.5 ?

between 0 and 1
f' = 2 - 2x
so
2 - 2x = 1.5
2x = .5
x = .25 is one spot

between 1 and 2
f' = 4 x - 4
so
1.5 = 4 x - 4
4 x = 5.5
x = 1.375 is another

Answer 2

Is f(x) continuous at x=1?
2x-x^2 = 1
2x^2-4x+3 = 1
So, yes.

The derivative exists everywhere in (0,2).

So, f(x) satisfies the MVT

f(2) = 3
f(0) = 0
So, the slope of the secant is is 3/2

Where does f'(c) = 3/2?
2-2x = 3/2
x = 1/4
y = 3/2 (x-1/4) + 7/16

4x-4 = 3/2
x = 11/8
y = 3/2 (x-11/8) + 41/32

See the graphs at

http://www.wolframalpha.com/input/?i=plot+y%3D2x-x^2%2C+y%3D2x^2-4x%2B3%2Cy+%3D+3x%2F2%2Cy+%3D+3%2F2+%28x-1%2F4%29+%2B+7%2F16%2Cy+%3D+3%2F2+%28x-11%2F8%29+%2B+41%2F32

it shows the parabolas, the secant, and the two tangents.