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!!!
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).
4 answers
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)
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!
1 answer