does the function satisfy the hypotheses of the Mean Value Theorem on the given interval?
f(x) = 2x^2 − 5x + 1, [0, 2]
If it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the Mean Value Theorem. (Enter your answers as a comma-separated list. If it does not satisfy the hypotheses, enter DNE).
Yes, f is continuous on [0, 2] and differentiable on (0, 2) since polynomials are continuous and differentiable on R.
I need the answer for second part.
Thanks!!!
just follow the theorem.
f(0) = 1
f(2) = -1
so, the average slope on the interval is (-1-1)/(2-0) = -1
So, now we need to find c such that f'(c) = -1
f'(x) = 4x-5
so, where does 4x-5 = -1? x = 1
1 is inside [0,2], as predicted by the Theorem.
In fact, since f(1) = -2, the line
y = -x-1
should be tangent to f(x) at x=1.
The graphs at
http://www.wolframalpha.com/input/?i=plot+y%3D2x^2%E2%88%925x+%2B+1%2C+y%3D-x-1%2C+y%3D-x%2B1
show that the tangent at x=1 is parallel to the line through f(0) and f(2)
THANK YOU SO MUCH FOR YOUR HELP STEVE!!!!
remember: wolframalpha is your friend!
To determine if the given function satisfies the hypotheses of the Mean Value Theorem (MVT) on the given interval [0, 2], we need to check two conditions:
1. f(x) must be continuous on the interval [0, 2].
2. f(x) must be differentiable on the open interval (0, 2).
First, let's check if f(x) = 2x^2 - 5x + 1 is continuous on the interval [0, 2].
To do this, we need to ensure that the function does not have any jumps, holes, or vertical asymptotes within the interval. In this case, the given function is a polynomial, and polynomials are continuous in their entire domain. Therefore, f(x) = 2x^2 - 5x + 1 is continuous on [0, 2].
Next, let's check if f(x) = 2x^2 - 5x + 1 is differentiable on the open interval (0, 2).
To do this, we need to make sure the function is smooth without any sharp corners or vertical tangents within the interval. Since f(x) is a polynomial, it is differentiable everywhere in its domain. Therefore, f(x) = 2x^2 - 5x + 1 is differentiable on (0, 2).
Since the function satisfies both conditions, it satisfies the hypotheses of the Mean Value Theorem on the given interval.
To find all the values of c that satisfy the conclusion of the Mean Value Theorem, we need to find the derivative of f(x), which will give us the slope of the secant line connecting the endpoints of the interval [0, 2]. Then we can find the point(s) on the function where the instantaneous rate of change (the derivative) equals the average rate of change (the slope of the secant line).
Taking the derivative of f(x), we get:
f'(x) = 4x - 5
Next, we calculate the average rate of change (slope of the secant line):
Average rate of change = [f(2) - f(0)] / [2 - 0] = [2(2)^2 - 5(2) + 1 - (2(0)^2 - 5(0) + 1)] / 2
= [8 - 10 + 1 - 1] / 2
= -1 / 2
To find the value(s) of c that satisfy the conclusion of the Mean Value Theorem, we need to find the value(s) of x in the open interval (0, 2) where f'(x) equals the average rate of change, -1/2.
Setting f'(x) equal to -1/2, we have:
4x - 5 = -1/2
Solving for x, we get:
4x = -1/2 + 5
4x = 9/2
x = 9/8
Therefore, the value of c that satisfies the conclusion of the Mean Value Theorem is c = 9/8.