Answer to Question #198598 in Calculus for desmond

Question

3. Suppose that f and g are continuous on [a, b] and differentiable on (a, b). Suppose also that f(a) = g(a) and f 0 (x) < g0 (x) for a < x < b. Prove that f(b) < g(b)

Answer 1

Actually Exact Question is :-

f and g are continuous on [a, b] and differentiable on (a, b). Suppose also that f(a) = g(a) and

f'(x) < g'(x) for a < x < b. Prove that f(b) < g(b).

Solution:-

Using the hint in the text look at the function h(x) = f(x) − g(x).

Note if h(b) < 0 then the

desired result follows.

Now apply the Mean Value Theorem to h.

Since f and g are continuous on [a, b]

and differentiable on (a, b) then so is h (the derivative is linear and the difference of continuous functions is

continuous).

The conditions of the Mean Value Theorem are satisfied for h so then there exists a c ∈ [a, b]

such that:

h(b) − h(a) = h'(c)· (b − a)

However recall that

h(a) = f(a) − g(a) = 0

since f(a) = g(a).

Also note that

h'(c) = f'(c) − g'(c) < 0

because, by assumption

f'(x) < g'(x) for all x.

Finally, because b > a, we have that

b − a > 0. Then:

h(b) = h(a) + h'(c) · (b − a)

= 0 + h'(c) · (b − a) < 0.

h(b)<0

Thus

f(b) − g(b) < 0,

so f(b) < g(b).