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
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
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.
To find two values of c in the interval [0,2] that satisfy the Mean Value Theorem, we need to first check if the given function is continuous on the interval [0,2] and differentiable on the open interval (0,2). If these conditions are met, then we can apply the Mean Value Theorem.
Let's start by checking if the function f(x) =
2x - x^2, if x < 1
2x^2 - 4x + 3, if x ≥ 1
is continuous on [0,2].
For a piecewise function to be continuous, the left-hand limit and right-hand limit at every point should be equal. In our case, we need to check the continuity at x = 1.
Taking the left-hand limit as x approaches 1:
lim(x->1-) 2x - x^2 = 2(1) - (1)^2 = 2 - 1 = 1
Taking the right-hand limit as x approaches 1:
lim(x->1+) 2x^2 - 4x + 3 = 2(1)^2 - 4(1) + 3 = 2 - 4 + 3 = 1
Since the left-hand limit and right-hand limit are both equal to 1, the function is continuous at x = 1.
Now, let's check if the function is differentiable on the open interval (0,2).
To be differentiable, the function must have a derivative at every point in the interval (0,2). We already know that the function has two different forms depending on whether x < 1 or x ≥ 1.
Differentiating the first part of the function:
f'(x) = 2 - 2x
Now, differentiating the second part of the function:
f'(x) = 4x - 4
Since both parts of the function have derivatives, the function is differentiable on the open interval (0,2).
Now that we have confirmed the function is continuous on [0,2] and differentiable on (0,2), we can apply the Mean Value Theorem.
According to the Mean Value Theorem, if a function is continuous on [a,b] and differentiable on (a,b), then there exists at least one value c in the interval (a,b) such that f'(c) = (f(b) - f(a))/(b - a).
In our case, a = 0 and b = 2.
Let's find the derivative of the function:
f'(x) =
2 - 2x, if x < 1
4x - 4, if x ≥ 1
Now we can use the Mean Value Theorem to find at least one value c.
(b - a) = (2 - 0) = 2
Let's find f(b) and f(a):
f(2) = 2(2) - (2)^2 = 4 - 4 = 0
f(0) = 2(0) - (0)^2 = 0 - 0 = 0
(f(b) - f(a))/(b - a) = (0 - 0)/2 = 0/2 = 0
The derivative of the function, f'(x), is the same for x < 1 and x ≥ 1, which means there is a value c in the interval (0,2) where f'(c) = 0. This satisfies the Mean Value Theorem.
To find another value c, we can try substituting different values in the interval (0,2) into the derivative equation and check if f'(c) is equal to 0. You can start with some values such as c = 0.5, c = 1.5, and so on, until you find another value that satisfies f'(c) = 0.
Keep in mind that since the function is different on each side of x = 1, there might be multiple values of c that satisfy the Mean Value Theorem in the given interval.