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Let fa n dg be differentiable on R and s...

Let `fa n dg` be differentiable on `R` and suppose `f(0)=g(0)a n df^(prime)(x)lt=g^(prime)(x)` for all `xgeq0.` Then show that `f(x)lt=g(x)` for all `xgeq0.`

Text Solution

Verified by Experts

Let h(x) =f(x)-g(x)
h(0)=f(0)=0
Now `h(x)-g(x)le0` for `xge0` (Given)
Thus h(x) is decreasing for `xgt0` Now
`xgt0`
`therefore h(x)leh(0)`
`therefore h(x)le0`
`therefore f(x)-g(x)le0`
`therefore f(x)leg(x)`
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