Question

Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval?

f(x) = 2x^2 − 3x + 1, [0, 2]

I found that c=3/4 but it was marked as wrong. Can you help?

Answer 1

it has to satisfy it because it is a polynomial and therefore continuous and differentiable everywhere

Answer 2

because it is a quadratic polynomial (degree 2) the number that satisfies the hypothesis of the MVT i.e. the \(c\) with \(f'(c)=\frac{f(b)-f(a)}{b-a}\) is actually the center of the interval

Answer 3

we can go through the work to find it if you like, but odds on it is 1

Answer 4

did you find \(f(2)\) ?

Answer 5

yes, f(2) =3

Answer 6

ok and \(f(0)=1\)

Answer 7

yes

Answer 8

so \(f(2)-f(0)=3-1=2\)

Answer 9

and therefore \(\frac{f(b)-f(a)}{b-a}=\frac{3-1}{2-0}=\frac{2}{2}=1\)

Answer 10

\[f'(x)=4x-3\] set
\[4x-3=1\]solve for \(x\) and i bet you get \(x=1\)

Answer 11

awesome. Thanks!

Answer 12

yw
any time you have a quadratic, the number you are looking for is always the center of the interval
you can prove it for yourself by considering a general quadratic
\[f(x)=ax^2+bx+c\]