Verify Mean Value Theorem, if $f(x)=x^{2}-4 x-3$ in the interval $[a, b],$ where $a=1$ and $b=4$
The given function is $f(x)=x^{2}-4 x-3$
$f,$ being a polynomial function, is a continuous in $[1,4]$ and is differentiable in $(1,4)$ whose derivative is $2 x-4$
$f(1)=1^{2}-4 \times 1-3=6, f(4)=4^{2}-4 \times 4-3=-3$
$\therefore \frac{f(b)-f(a)}{b-a}=\frac{f(4)-f(1)}{4-1}=\frac{-3-(-6)}{3}=\frac{3}{3}=1$
Mean Value Theorem states that there is a point $c \in(1,4)$ such that
$f^{\prime}(c)=1$ $f^{\prime}(c)=1$
$\Rightarrow 2 c-4=1$
$\Rightarrow c=\frac{5}{2},$ where $c=\frac{5}{2} \in(1,4)$
Hence, Mean Value Theorem is verified foer the given function.
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$f(x)=[x]$ for $x \in[-2,2]$
Let $\mathrm{f}$ be any continuous function on $[0,2]$ and twice differentiable on $(0,2)$. If $\mathrm{f}(0)=0, \mathrm{f}(1)=1$ and $f(2)=2$, then
Let $f$ and $g$ be twice differentiable even functions on $(-2,2)$ such that $f\left(\frac{1}{4}\right)=0, f\left(\frac{1}{2}\right)=0, f(1)=1$ and $g\left(\frac{3}{4}\right)=0, g(1)=2$ Then, the minimum number of solutions of $f(x) g^{\prime \prime}(x)+f^{\prime}(x) g^{\prime}(x)=0$ in $(-2,2)$ is equal to
Which of the following function can satisfy Rolle's theorem ?
If the Rolle's theorem holds for the function $f(x) = 2x^3 + ax^2 + bx$ in the interval $[-1, 1 ]$ for the point $c = \frac{1}{2}$ , then the value of $2a + b$ is