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.
Let $f(x)$ be a function continuous on $[1,2]$ and differentiable on $(1,2)$ satisfying
$f(1) = 2, f(2) = 3$ and $f'(x) \geq 1 \forall x \in (1,2)$.Define $g(x)=\int\limits_1^x {f(t)\,dt\,\forall \,x\, \in [1,2]} $ then the greatest value of $g(x)$ on $[1,2]$ is-
If $c$ is a point at which Rolle's theorem holds for the function, $f(\mathrm{x})=\log _{\mathrm{e}}\left(\frac{\mathrm{x}^{2}+\alpha}{7 \mathrm{x}}\right)$ in the interval $[3,4],$ where $\alpha \in \mathrm{R},$ then $f^{\prime \prime}(\mathrm{c})$ is equal to
Let $g: R \rightarrow R$ be a non constant twice differentiable such that $g^{\prime}\left(\frac{1}{2}\right)=g^{\prime}\left(\frac{3}{2}\right)$. If a real valued function $f$ is defined as $\mathrm{f}(\mathrm{x})=\frac{1}{2}[\mathrm{~g}(\mathrm{x})+\mathrm{g}(2-\mathrm{x})]$, then
If the function $f(x) = {x^3} - 6a{x^2} + 5x$ satisfies the conditions of Lagrange's mean value theorem for the interval $[1, 2] $ and the tangent to the curve $y = f(x) $ at $x = {7 \over 4}$ is parallel to the chord that joins the points of intersection of the curve with the ordinates $x = 1$ and $x = 2$. Then the value of $a$ is
The function $f(x) = x(x + 3){e^{ - (1/2)x}}$ satisfies all the conditions of Rolle's theorem in $ [-3, 0]$. The value of $c$ is