Question

Mean Value Theorem Help!!!!!!!

Answer 1

Suppose that \[3\le f'(x) \le5\] for all values of x. Show that \[18\le f(8)-f(2) \le30\]

Answer 2

Consider the integral \[\huge \int_2^8 f'(x) \ dx\]
This is of course equal to
\[\huge \int_2^8 f'(x) \ dx= f(8) - f(2)\]

Now, since the least value taken on by \(f'(x)\) is 3 the smallest that integral could be would be the length of the integral times 3, or 6*3 = 18. Do you see how 18 is the lower bound? If yes, the same reasoning shows how 30 is the upper bound.

References: http://tutorial.math.lamar.edu/Classes/CalcI/DefnofDefiniteIntegral.aspx
after example 5 you'll see a list of properties, # 10 is closely related to your problem.

Answer 3

we haven't learned integrals yet

Answer 4

The MVT tells us that when its conditions are satisfied, there is some c in the range between a and b such that\[\frac{ f(b)-f(a) }{ b-a }=f'(c)\]In this case we want to multiply through by (b-a):\[f(b)-f(a)=(b-a)f'(c)\]This translates into\[f(8)-f(2)=6 \times f'(c)\]We don't know exactly what f'(c) is, but we're told that f'(x) is always in the range from 3 to 5, and that establishes the inequality in this problem.