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Question

(D) Define a function $h(x) = f(x) - x$.Given $f(\frac{1}{2}) = \frac{1}{2}$ and $f(1) = 1$,we have $h(\frac{1}{2}) = f(\frac{1}{2}) - \frac{1}{2} = 0

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$f^{\prime}(1) > 1$

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(D) Define a function $h(x) = f(x) - x$.Given $f(\frac{1}{2}) = \frac{1}{2}$ and $f(1) = 1$,we have $h(\frac{1}{2}) = f(\frac{1}{2}) - \frac{1}{2} = 0$ and $h(1) = f(1) - 1 = 0$.Since $f(x)$ is twice differentiable,$h(x)$ is also differentiable on $[\frac{1}{2}, 1]$.By Rolle's Theorem,there exists some $\alpha \in (\frac{1}{2}, 1)$ such that $h^{\prime}(\alpha) = 0$.Since $h^{\prime}(x) = f^{\prime}(x) - 1$,this implies $f^{\prime}(\alpha) = 1$.We are given $f^{\prime \prime}(x) > 0$ for all $x \in R$,which means $f^{\prime}(x)$ is a strictly increasing function.Since $\alpha Substituting $f^{\prime}(\alpha) = 1$,we get $1 < f^{\prime}(1)$.

If $f: R \rightarrow R$ is a twice differentiable function such that $f^{\prime \prime}(x) > 0$ for all $x \in R$,and $f(\frac{1}{2}) = \frac{1}{2}$,$f(1) = 1$,then

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