Suppose that f(0) = −5 and f '(x) ≤ 7 for all values of x. how large can f(5) possibly be?

f(5) cannot exceed 30.

Step-by-step explanation:

f is differentiable in all R, so we can apply the Mean Value Theorem on the interval [0,5], since f is continuous on [0,5] and differentiable on (0,5).

By the MVT, there exists some number c∈(0,5) such that . Substituting the value of f(0), this last equation is equivalent to . But f'(x)≤7 for all values of x, in particular for x=c, so applying this bound we obtain that . Then , so substracting 5 from both sides we conclude that .

idk i would have to be there to under stand it

step-by-step explanation:

idk i would have to be there to under stand it

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!