Verify Lagrange's mean value theorem for the following

Solution:

(i) We have

$f(x)=x^{2}-1$

Since a polynomial function is everywhere continuous and differentiable, $f(x)$ is continuous on $[2,3]$ and differentiable on $(2,3)$. Thus, both conditions of Lagrange's mean value theorem are satisfied. So, there must exist at least one real number $c \in(2,3)$ such that

$f^{\prime}(c)=\frac{f(3)-f(2)}{3-2}$

Now. $f(x)=x^{2}-1$

$\Rightarrow f^{\prime}(x)=2 x, f(3)=(3)^{2}-1=8, f(2)=(2)^{2}-1=3$

$\therefore f^{\prime}(x)=\frac{f(3)-f(2)}{3-2}$

$\Rightarrow 2 x=\frac{8-3}{1}$

$\Rightarrow x=\frac{5}{2}$

Thus, $c=\frac{5}{2} \in(2,3)$ such that $f^{\prime}(c)=\frac{f(3)-f(2)}{3-2}$.

Hence, Lagrange's theorem is verified.

(ii) We have,

$f(x)=x^{3}-2 x^{2}-x+3=0$

Since a polynomial function is everywhere continuous and differentiable

Therefore, $f(x)$ is continuous on $[0,1]$ and differentiable on $(0,1)$.

Thus, both conditions of Lagrange's mean value theorem are satisfied. So, there must exist at least one real number $c \in(0,1)$ such that

$f^{\prime}(c)=\frac{f(1)-f(0)}{1-0}$

Now, $f(x)=x^{3}-2 x^{2}-x+3=0$

$\Rightarrow f^{\prime}(x)=3 x^{2}-4 x-1, f(1)=1, f(0)=3$

$\therefore f^{\prime}(x)=\frac{f(1)-f(0)}{1-0}$

$\Rightarrow 3 x^{2}-4 x-1=\frac{1-3}{1}$

$\Rightarrow 3 x^{2}-4 x-1+2=0$

$\Rightarrow 3 x^{2}-4 x+1=0$

$\Rightarrow 3 x^{2}-3 x-x+1=0$

$\Rightarrow(3 x-1)(x-1)=0$

$\Rightarrow x=\frac{1}{3}, 1$

Thus, $c=\frac{1}{3} \in(0,1)$ such that $f^{\prime}(c)=\frac{f(1)-f(0)}{1-0}$.

Hence, Lagrange's theorem is verified.

(iii) We have,

$f(x)=x(x-1)$ which can be rewritten as $f(x)=x^{2}-x$

Since a polynomial function is everywhere continuous and differentiable.

Therefore, $f(x)$ is continuous on $[1,2]$ and differentiable on $(1,2)$.

Thus, both conditions of Lagrange's mean value theorem are satisfied. So, there must exist at least one real number $c \in(1,2)$ such that

$f^{\prime}(c)=\frac{f(2)-f(1)}{2-1}$

Now, $f(x)=x^{2}-x$

$\Rightarrow f^{\prime}(x)=2 x-1, f(2)=2, f(1)=0$

$\therefore f^{\prime}(x)=\frac{f(2)-f(1)}{2-1}$

$\Rightarrow 2 x-1=\frac{2-0}{2-1}$

$\Rightarrow 2 x-1-2=0$

$\Rightarrow 2 x=3$

$\Rightarrow x=\frac{3}{2}$

Thus, $c=\frac{3}{2} \in(1,2)$ such that $f^{\prime}(c)=\frac{f(2)-f(1)}{2-1}$.

Hence, Lagrange's theorem is verified.

(iv) We have,

$f(x)=x^{2}-3 x+2$

Since a polynomial function is everywhere continuous and differentiable.

Therefore, $f(x)$ is continuous on $[-1,2]$ and differentiable on $(-1,2)$. Thus, both conditions of Lagrange's mean value theorem are satisfied. So, there must exist at least one real number $c \in(-1,2)$ such that

$f^{\prime}(c)=\frac{f(2)-f(-1)}{2+1}=\frac{f(2)-f(-1)}{3}$

Now, $f(x)=x^{2}-3 x+2$

$\Rightarrow f^{\prime}(x)=2 x-3, f(2)=0, f(-1)=(-1)^{2}-3(-1)+2=6$

$\therefore f^{\prime}(x)=\frac{f(2)-f(-1)}{3}$

$\Rightarrow 2 x-3=-2$

$\Rightarrow 2 x-1=0$

$\Rightarrow x=\frac{1}{2}$

Thus, $c=\frac{1}{2} \in(-1,2)$ such that $f^{\prime}(c)=\frac{f(2)-f(-1)}{2-(-1)}$.

Hence, Lagrange's theorem is verified.

(v) We have,

$f(x)=2 x^{2}-3 x+1$

Since a polynomial function is everywhere continuous and differentiable.

Therefore, $f(x)$ is continuous on $[1,3]$ and differentiable on $(1,3)$. Thus, both conditions of Lagrange's mean value theorem are satisfied. So, there must exist at least one real number $c \in(1,3)$ such that

$f^{\prime}(c)=\frac{f(3)-f(1)}{3-1}=\frac{f(3)-f(1)}{2}$

Now, $f(x)=2 x^{2}-3 x+1$

$\Rightarrow f^{\prime}(x)=4 x-3, f(3)=10, f(1)=2(1)^{2}-3(1)+1=0$

$\therefore f^{\prime}(x)=\frac{f(3)-f(1)}{2}$

$\Rightarrow 4 x-3=\frac{10-0}{2}$

$\Rightarrow 4 x-3-5=0$

$\Rightarrow x=2$

Thus, $c=2 \in(1,3)$ such that $f^{\prime}(c)=\frac{f(3)-f(1)}{3-1}$.

Hence, Lagrange's theorem is verified.

(vi) We have,

$f(x)=x^{2}-2 x+4$

Since a polynomial function is everywhere continuous and differentiable.

Therefore, $f(x)$ is continuous on $[1,5]$ and differentiable on $(1,5)$. Thus, both conditions of Lagrange's mean value theorem are satisfied. So, there must exist at least one real number $c \in(1,5)$ such that

$f^{\prime}(c)=\frac{f(5)-f(1)}{5-1}=\frac{f(5)-f(1)}{4}$

Now, $f(x)=x^{2}-2 x+4$

$\Rightarrow f^{\prime}(x)=2 x-2, f(5)=25-10+4=19, f(1)=1-2+4=3$

$\therefore f^{\prime}(x)=\frac{f(5)-f(1)}{4}$

$\Rightarrow 2 x-2=\frac{19-3}{4}$

$\Rightarrow 2 x-2-4=0$

$\Rightarrow x=\frac{6}{2}=3$

Thus, $c=3 \in(1,5)$ such that $f^{\prime}(c)=\frac{f(5)-f(1)}{5-1}$.

Hence, Lagrange's theorem is verified.

(vii) We have,

$f(x)=2 x-x^{2}$

Since a polynomial function is everywhere continuous and differentiable.

Therefore, $f(x)$ is continuous on $[0,1]$ and differentiable on $(0,1)$. Thus, both conditions of Lagrange's mean value theorem are satisfied. So, there must exist at least one real number $c \in(0,1)$ such that

$f^{\prime}(c)=\frac{f(1)-f(0)}{1-0}=\frac{f(1)-f(0)}{1}$

Now, $f(x)=2 x-x^{2}$

$\Rightarrow f^{\prime}(x)=2-2 x, f(1)=2-1=1, f(0)=0$

$\therefore f^{\prime}(x)=\frac{f(1)-f(0)}{1}$

$\Rightarrow 2-2 x=\frac{1-0}{1}$

$\Rightarrow-2 x=1-2$

$\Rightarrow x=\frac{1}{2}$

Thus, $c=\frac{1}{2} \in(0,1)$ such that $f^{\prime}(c)=\frac{f(1)-f(0)}{1-0}$.

Hence, Lagrange's theorem is verified.

(viii) We have,

$f(x)=(x-1)(x-2)(x-3)$ which can be rewritten as $f(x)=x^{3}-6 x^{2}+11 x-6$

Since a polynomial function is everywhere continuous and differentiable.

Therefore, $f(x)$ is continuous on $[0,4]$ and differentiable on $(0,4)$. Thus, both conditions of Lagrange's mean value theorem are satisfied. So, there must exist at least one real number $c \in(0,4)$ such that

$f^{\prime}(c)=\frac{f(4)-f(0)}{4-0}=\frac{f(4)-f(0)}{4}$

Now, $f(x)=x^{3}-6 x^{2}+11 x-6$

$\Rightarrow f^{\prime}(x)=3 x^{2}-12 x+11, f(0)=-6, f(4)=64-96+44-6=6$

$\therefore f^{\prime}(x)=\frac{f(4)-f(0)}{4-0}$

$\Rightarrow 3 x^{2}-12 x+11=\frac{6+6}{4}$

$\Rightarrow 3 x^{2}-12 x+8=0$

$\Rightarrow x=2-\frac{2}{\sqrt{3}}, 2+\frac{2}{\sqrt{3}}$

Thus, $c=2 \pm \frac{2}{\sqrt{3}} \in(0,4)$ such that $f^{\prime}(c)=\frac{f(4)-f(0)}{4-0}$.

Hence, Lagrange's theorem is verified.

(ix) We have,

$f(x)=\sqrt{25-x^{2}}$

Here, $f(x)$ will exist,

if

$25-x^{2} \geq 0$

$\Rightarrow x^{2} \leq 25$

$\Rightarrow-5 \leq x \leq 5$

Since for each $x \in[-3,4]$, the function $f(x)$ attains a unique definite value.

So, $f(x)$ is continuous on $[-3,4]$

Also, $f^{\prime}(x)=\frac{1}{2 \sqrt{25-x^{2}}}(-2 x)=\frac{-x}{\sqrt{25-x^{2}}}$ exists for all $x \in(-3,4)$

So, $f(x)$ is differentiable on $(-3,4)$.

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some $c \in(-3,4)$ such that

$f^{\prime}(c)=\frac{f(4)-f(-3)}{4+3}=\frac{f(4)-f(-3)}{7}$

Now, $f(x)=\sqrt{25-x^{2}}$

$f^{\prime}(x)=\frac{-x}{\sqrt{25-x^{2}}}, f(-3)=4, f(4)=3$

$\therefore f^{\prime}(x)=\frac{f(4)-f(-3)}{4+3}$

$\Rightarrow \frac{-x}{\sqrt{25-x^{2}}}=\frac{3-4}{7}$

$\Rightarrow 49 x^{2}=25-x^{2}$

$\Rightarrow x=\pm \frac{1}{\sqrt{2}}$

Thus, $c=\pm \frac{1}{\sqrt{2}} \in(-3,4)$ such that $f^{\prime}(c)=\frac{f(4)-f(-3)}{4-(-3)}$.

Hence, Lagrange's theorem is verified.

(x) We have,

$f(x)=\tan ^{-1} x$

Clearly, $f(x)$ is continuous on $[0,1]$ and derivable on $(0,1)$

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some $c \in(-3,4)$ such that

$f^{\prime}(c)=\frac{f(1)-f(0)}{1-0}=\frac{f(1)-f(0)}{1}$

$f^{\prime}(c)=\frac{f(1)-f(0)}{1-0}=\frac{f(1)-f(0)}{1}$

$\therefore f^{\prime}(x)=\frac{f(1)-f(0)}{1-0}$

$\Rightarrow \frac{1}{1+x^{2}}=\frac{\pi}{4}-0$

$\Rightarrow\left(\frac{4}{\pi}-1\right)=x^{2}$

$\Rightarrow x=\pm \sqrt{\frac{4-\pi}{\pi}}$

Thus, $c=\sqrt{\frac{4-\pi}{\pi}} \in(0,1)$ such that $f^{\prime}(c)=\frac{f(1)-f(0)}{1-0}$.

Hence, Lagrange's theorem is verified.

(xi) We have,

Clearly, $f(x)$ is continuous on $[1,3]$ and derivable on $(1,3)$

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some $c \in(1,3)$ such that

$f^{\prime}(c)=\frac{f(3)-f(1)}{3-1}=\frac{f(3)-f(1)}{2}$

Now, $f(x)=\frac{x^{2}+1}{x}$

$f^{\prime}(x)=\frac{x^{2}-1}{x^{2}}, f(1)=2, f(3)=\frac{10}{3}$

$\therefore f^{\prime}(x)=\frac{f(3)-f(1)}{2}$

$\Rightarrow \frac{x^{2}-1}{x^{2}}=\frac{4}{6}$

$\Rightarrow \frac{x^{2}-1}{x^{2}}=\frac{2}{3}$

$\Rightarrow 3 x^{2}-3=2 x^{2}$

$\Rightarrow x=\pm \sqrt{3}$

Thus, $c=\sqrt{3} \in(1,3)$ such that $f^{\prime}(c)=\frac{f(3)-f(1)}{3-1}$.

Hence, Lagrange's theorem is verified.

(xii) We have,

$f(x)=x(x+4)^{2}=x\left(x^{2}+16+8 x\right)=x^{3}+8 x^{2}+16 x$

Since $f(x)$ is a polynomial function which is everywhere continuous and differentiable.

Therefore, $f(x)$ is continuous on $[0,4]$ and derivable on $(0,4)$

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some $c \in(0,4)$ such that

$f^{\prime}(c)=\frac{f(4)-f(0)}{4-0}=\frac{f(4)-f(0)}{4}$

Now, $f(x)=x^{3}+8 x^{2}+16 x$

$f^{\prime}(x)=3 x^{2}+16 x+16, f(4)=64+128+64=256, f(0)=0$

$\therefore f^{\prime}(x)=\frac{f(4)-f(0)}{4-0}$

$\Rightarrow 3 x^{2}+16 x+16=\frac{256}{4}$

$\Rightarrow 3 x^{2}+16 x-48=0$

$\Rightarrow x=-\frac{4}{3}(2+\sqrt{13}), \frac{4}{3}(\sqrt{13}-2)$

Thus, $c=\frac{-8+4 \sqrt{13}}{3} \in(0,4)$ such that $f^{\prime}(c)=\frac{f(4)-f(0)}{4-0}$.

Hence, Lagrange's theorem is verified.

(xiii) We have,

$f(x)=\sqrt{x^{2}-4}$

Here, $f(x)$ will exist,

if

$x^{2}-4 \geq 0$

$\Rightarrow x \leq-2$ or $x \geq 2$

Since for each $x \in[2,4]$, the function $f(x)$ attains a unique definite value.

So, $f(x)$ is continuous on $[2,4]$

Also, $f^{\prime}(x)=\frac{1}{2 \sqrt{x^{2}-4}}(2 x)=\frac{x}{\sqrt{x^{2}-4}}$ exists for all $x \in(2,4)$

So, $f(x)$ is differentiable on $(2,4)$.

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some $c \in(2,4)$ such that

$f^{\prime}(c)=\frac{f(4)-f(2)}{4-2}=\frac{f(4)-f(2)}{2}$

$\therefore f^{\prime}(x)=\frac{f(4)-f(2)}{4-2}$

$\Rightarrow \frac{x}{\sqrt{x^{2}-4}}=\frac{2 \sqrt{3}}{2}$

$\Rightarrow \frac{x}{\sqrt{x^{2}-4}}=\sqrt{3}$

$\Rightarrow \frac{x^{2}}{x^{2}-4}=3$

$\Rightarrow x^{2}=3 x^{2}-12$

$\Rightarrow x^{2}=6$

$\Rightarrow x=\pm \sqrt{6}$

Thus, $c=\sqrt{6} \in(2,4)$ such that $f^{\prime}(c)=\frac{f(4)-f(2)}{4-2}$.

Hence, Lagrange's theorem is verified.

(xiv) We have,

$f(x)=x^{2}+x-1$

Since polynomial function is everywhere continuous and differentiable.

Therefore, $f(x)$ is continuous on $[0,4]$ and differentiable on $(0,4)$

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some $c \in(0,4)$ such that

$f^{\prime}(c)=\frac{f(4)-f(0)}{4-0}=\frac{f(4)-f(0)}{4}$

$\therefore f^{\prime}(x)=\frac{f(4)-f(0)}{4-0}$

$\Rightarrow 2 x+1=\frac{20}{4}$

$\Rightarrow 2 x+1=5$

$\Rightarrow 2 x=4$

$\Rightarrow x=2$

Thus, $c=2 \in(0,4)$ such that $f^{\prime}(c)=\frac{f(4)-f(0)}{4-0}$.

Hence, Lagrange's theorem is verified

(xv) We have,

$f(x)=\sin x-\sin 2 x-x$

Since $\sin x, \sin 2 x \& x$ are everywhere continuous and differentiable

Therefore, $f(x)$ is continuous on $[0, \pi]$ and differentiable on $(0, \pi)$

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some $c \in(0, \pi)$ such that

$f^{\prime}(c)=\frac{f(\pi)-f(0)}{\pi-0}=\frac{f(\pi)-f(0)}{\pi}$

$\therefore f^{\prime}(x)=\frac{f(\pi)-f(0)}{\pi-0}$

$\Rightarrow \cos x-2 \cos 2 x-1=-1$

$\Rightarrow \cos x-2 \cos 2 x=0$

$\Rightarrow \cos x-4 \cos ^{2} x=-2$

$\Rightarrow 4 \cos ^{2} x-\cos x-2=0$

$\Rightarrow \cos x=\frac{1}{8}(1 \pm \sqrt{33})$

$\Rightarrow x=\cos ^{-1}\left[\frac{1}{8}(1 \pm \sqrt{33})\right]$

Thus, $c=\cos ^{-1}\left(\frac{1 \pm \sqrt{33}}{8}\right) \in(0, \pi)$ such that $f^{\prime}(c)=\frac{f(\pi)-f(0)}{\pi-0}$.

Hence, Lagrange's theorem is verified.'

(xvi) We have,

$f(x)=x^{3}-5 x^{2}-3 x$

Since polynomial function is everywhere continuous and differentiable

Therefore, $f(x)$ is continuous on $[1,3]$ and differentiable on $(1,3)$

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some $c \in(1,3)$ such that

$f^{\prime}(c)=\frac{f(3)-f(1)}{3-1}=\frac{f(3)-f(1)}{2}$

Now, $f(x)=x^{3}-5 x^{2}-3 x$

$f^{\prime}(x)=3 x^{2}-10 x-3, f(3)=-27, f(1)=-7$

$\therefore f^{\prime}(x)=\frac{f(3)-f(1)}{2}$

$\Rightarrow 3 x^{2}-10 x-3=\frac{-20}{2}$

$\Rightarrow 3 x^{2}-10 x+7=0$

$\Rightarrow x=1, \frac{7}{3}$

Thus, $c=\frac{7}{3} \in(1,3)$ such that $f^{\prime}(c)=\frac{f(3)-f(1)}{3-1}$.

Hence, Lagrange's theorem is verified.