Medium
If $f:[-5,5] \rightarrow R$ is a differentiable function and if $f^{\prime}(x)$ does not vanish anywhere,then prove that $f(-5) \neq f(5).$
(N/A) It is given that $f:[-5,5] \rightarrow R$ is a differentiable function.
Since every differentiable function is a continuous function,we obtain:
$a) f$ is continuous on $[-5,5].$
$b) f$ is differentiable on $(-5,5).$
Therefore,by the Mean Value Theorem,there exists $c \in (-5,5)$ such that:
$f^{\prime}(c) = \frac{f(5) - f(-5)}{5 - (-5)}$
$\Rightarrow 10 f^{\prime}(c) = f(5) - f(-5)$
It is also given that $f^{\prime}(x)$ does not vanish anywhere,which means $f^{\prime}(x) \neq 0$ for all $x \in [-5,5].$
Therefore,$f^{\prime}(c) \neq 0.$
$\Rightarrow 10 f^{\prime}(c) \neq 0$
$\Rightarrow f(5) - f(-5) \neq 0$
$\Rightarrow f(5) \neq f(-5)$
Hence,it is proved that $f(-5) \neq f(5).$
Since every differentiable function is a continuous function,we obtain:
$a) f$ is continuous on $[-5,5].$
$b) f$ is differentiable on $(-5,5).$
Therefore,by the Mean Value Theorem,there exists $c \in (-5,5)$ such that:
$f^{\prime}(c) = \frac{f(5) - f(-5)}{5 - (-5)}$
$\Rightarrow 10 f^{\prime}(c) = f(5) - f(-5)$
It is also given that $f^{\prime}(x)$ does not vanish anywhere,which means $f^{\prime}(x) \neq 0$ for all $x \in [-5,5].$
Therefore,$f^{\prime}(c) \neq 0.$
$\Rightarrow 10 f^{\prime}(c) \neq 0$
$\Rightarrow f(5) - f(-5) \neq 0$
$\Rightarrow f(5) \neq f(-5)$
Hence,it is proved that $f(-5) \neq f(5).$
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Let $f(x)$ and $g(x)$ be two differentiable functions in $R$ such that $f(2) = 8, g(2) = 0, f(4) = 10$,and $g(4) = 8$. Then which of the following is true?
Let $f(x)$ be a non-constant twice differentiable function defined on $(-\infty, \infty)$ such that $f(x)=f(1-x)$ and $f^{\prime}\left(\frac{1}{4}\right)=0$. Then
$(A)$ $f^{\prime \prime}(x)$ vanishes at least twice on $[0,1]$
$(B)$ $f^{\prime}\left(\frac{1}{2}\right)=0$
$(C)$ $\int_{-1 / 2}^{1 / 2} f\left(x+\frac{1}{2}\right) \sin x d x=0$
$(D)$ $\int_0^{1 / 2} f(t) e^{\sin \pi t} d t=\int_{1 / 2}^1 f(1-t) e^{\sin \pi t} d t$
$(A)$ $f^{\prime \prime}(x)$ vanishes at least twice on $[0,1]$
$(B)$ $f^{\prime}\left(\frac{1}{2}\right)=0$
$(C)$ $\int_{-1 / 2}^{1 / 2} f\left(x+\frac{1}{2}\right) \sin x d x=0$
$(D)$ $\int_0^{1 / 2} f(t) e^{\sin \pi t} d t=\int_{1 / 2}^1 f(1-t) e^{\sin \pi t} d t$
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Difficult
View SolutionLet $f: R \rightarrow R$ be a twice continuously differentiable function such that $f(0) = f(1) = f^{\prime}(0) = 0$. Then:
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