Let f(x) and g(x) be differentiable functions on R and f(0) = g(0) and f′(x) ≤ g′(x) for all x. Prove that if?
a > 0, then f(a) ≤ g(a). (There is a solution that uses the Mean Value Theorem)
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a > 0, then f(a) ≤ g(a). (There is a solution that uses the Mean Value Theorem)
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Verified answer
Here is the solution using MVT.
Let h(x)=g(x)-f(x). Clearly h is differentiable on (0,a) and continuous on [0,a]. So by MVT
h(a)-h(0) = h'(c) (a-0)
Since h(0)=g(0)-f(0)=0 and h'(c)=g'(c)-f'(c)>=0
h(a) =h'(c) *a >= 0
So g(a)-f(a) >=0, or equivalently g(a)>=f(a).