Question

Verify Lagrange's mean value theorem for the following functions on the indicated intervals. In each case find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem
(i) f(x) = x² − 1 on [2, 3]
(ii) f(x) = x³ − 2x² − x + 3 on [0, 1]
(iii) f(x) = x(x −1) on [1, 2]
(iv) f(x) = x² − 3x + 2 on [−1, 2]
(v) f(x) = 2x² − 3x + 1 on [1, 3]
(vi) f(x) = x² − 2x + 4 on [1, 5]
(vii) f(x) = 2x − x² on [0, 1]
(viii) f(x) = (x − 1)(x − 2)(x − 3) on [0, 4]
(ix) $f\left(x\right)=\sqrt{25-x^2}$ on [−3, 4]
(x) f(x) = tan⁻¹ x on [0, 1]
(xi) $f\left(x\right)=x+\frac{1}{x}\mathrm{on}[1,3]$
(xii) f(x) = x(x + 4)² on [0, 4]
(xiii) $f\left(x\right)=\sqrt{x^2-4}\mathrm{on}[2,4]$
(xiv) f(x) = x² + x − 1 on [0, 4]
(xv) f(x) = sin x − sin 2x − x on [0, π]
(xvi) f(x) = x³ − 5x² − 3x on [1, 3]

Answer 1

(i) We have

$f\left(x\right)=x^2-1$

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

Now, $f\left(x\right)=x^2-1$
$\Rightarrow f\text{'}\left(x\right)=2x$ , $f\left(3\right)={\left(3\right)}^2-1=8$ , $f\left(2\right)={\left(2\right)}^2-1=3$

∴ $f\text{'}\left(x\right)=\frac{f\left(3\right)-f\left(2\right)}{3-2}$

$$\begin{gathered} \Rightarrow 2x=\frac{8-3}{1} \\ \Rightarrow x=\frac{5}{2} \end{gathered}$$
Thus, $c=\frac{5}{2}\in \left(2,3\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(2\right)}{3-2}$.

Hence, Lagrange's theorem is verified.

(ii) We have,

$f\left(x\right)=x^3-2x^2-x+3=0$

Since a polynomial function is everywhere continuous and differentiable.

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

Now, $f\left(x\right)=x^3-2x^2-x+3=0$
$\Rightarrow f\text{'}\left(x\right)=3x^2-4x-1$ , $f\left(1\right)=1$ , $f\left(0\right)=3$

∴ $f\text{'}\left(x\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$

$$\begin{gathered} \Rightarrow 3x^2-4x-1=\frac{1-3}{1} \\ \Rightarrow 3x^2-4x-1+2=0 \\ \Rightarrow 3x^2-4x+1=0 \\ \Rightarrow 3x^2-3x-x+1=0 \\ \Rightarrow \left(3x-1\right)\left(x-1\right)=0 \\ \Rightarrow x=\frac{1}{3},1 \end{gathered}$$
Thus, $c=\frac{1}{3}\in \left(0,1\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$.

Hence, Lagrange's theorem is verified.

(iii) We have,

$f\left(x\right)=x\left(x-1\right)$ which can be rewritten as $f\left(x\right)=x^2-x$

Since a polynomial function is everywhere continuous and differentiable.

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

Now, $f\left(x\right)=x^2-x$

$\Rightarrow f\text{'}\left(x\right)=2x-1$ , $f\left(2\right)=2$ , $f\left(1\right)=0$

∴ $f\text{'}\left(x\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}$

$$\begin{gathered} \Rightarrow 2x-1=\frac{2-0}{2-1} \\ \Rightarrow 2x-1-2=0 \\ \Rightarrow 2x=3 \\ \Rightarrow x=\frac{3}{2} \end{gathered}$$
Thus, $c=\frac{3}{2}\in \left(1,2\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}$.

Hence, Lagrange's theorem is verified.

(iv) We have,

$f\left(x\right)=x^2-3x+2$

Since a polynomial function is everywhere continuous and differentiable.

Therefore, $f\left(x\right)$ is continuous on $\left[-1,2\right]$ and differentiable on $\left(-1,2\right)$.
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number ​$c\in \left(-1,2\right)$ such that
$f\text{'}\left(c\right)=\frac{f\left(2\right)-f\left(-1\right)}{2+1}=\frac{f\left(2\right)-f\left(-1\right)}{3}$

Now, $f\left(x\right)=x^2-3x+2$
$\Rightarrow f\text{'}\left(x\right)=2x-3$ , $f\left(2\right)=0$ , $f\left(-1\right)={\left(-1\right)}^2-3\left(-1\right)+2=6$

∴ $f\text{'}\left(x\right)=\frac{f\left(2\right)-f\left(-1\right)}{3}$

$$\begin{gathered} \Rightarrow 2x-3=-2 \\ \Rightarrow 2x-1=0 \\ \Rightarrow x=\frac{1}{2} \end{gathered}$$
Thus, $c=\frac{1}{2}\in \left(-1,2\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(2\right)-f\left(-1\right)}{2-\left(-1\right)}$.

Hence, Lagrange's theorem is verified.

(v) We have,

$f\left(x\right)=2x^2-3x+1$

Since a polynomial function is everywhere continuous and differentiable.

Therefore, $f\left(x\right)$ is continuous on $\left[1,3\right]$ and differentiable on $\left(1,3\right)$.
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number ​$c\in \left(1,3\right)$ such that
$f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}=\frac{f\left(3\right)-f\left(1\right)}{2}$

Now, $f\left(x\right)=2x^2-3x+1$
$\Rightarrow f\text{'}\left(x\right)=4x-3$ , $f\left(3\right)=10$ , $f\left(1\right)=2{\left(1\right)}^2-3\left(1\right)+1=0$

∴ $f\text{'}\left(x\right)=\frac{f\left(3\right)-f\left(1\right)}{2}$

$$\begin{gathered} \Rightarrow 4x-3=\frac{10-0}{2} \\ \Rightarrow 4x-3-5=0 \\ \Rightarrow x=2 \end{gathered}$$
Thus, $c=2\in \left(1,3\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}$.

Hence, Lagrange's theorem is verified.

(vi) We have,

$f\left(x\right)=x^2-2x+4$

Since a polynomial function is everywhere continuous and differentiable.

Therefore, $f\left(x\right)$ is continuous on $\left[1,5\right]$ and differentiable on $\left(1,5\right)$.
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number ​$c\in \left(1,5\right)$ such that
$f\text{'}\left(c\right)=\frac{f\left(5\right)-f\left(1\right)}{5-1}=\frac{f\left(5\right)-f\left(1\right)}{4}$

Now, $f\left(x\right)=x^2-2x+4$
$\Rightarrow f\text{'}\left(x\right)=2x-2$ , $f\left(5\right)=25-10+4=19$ , $f\left(1\right)=1-2+4=3$

∴ $f\text{'}\left(x\right)=\frac{f\left(5\right)-f\left(1\right)}{4}$

$$\begin{gathered} \Rightarrow 2x-2=\frac{19-3}{4} \\ \Rightarrow 2x-2-4=0 \\ \Rightarrow x=\frac{6}{2}=3 \end{gathered}$$
Thus, $c=3\in \left(1,5\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(5\right)-f\left(1\right)}{5-1}$.

Hence, Lagrange's theorem is verified.

(vii) We have,

$f\left(x\right)=2x-x^2$

Since a polynomial function is everywhere continuous and differentiable.

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

Now, $f\left(x\right)=2x-x^2$
$\Rightarrow f\text{'}\left(x\right)=2-2x$ , $f\left(1\right)=2-1=1$ , $f\left(0\right)=0$

∴ $f\text{'}\left(x\right)=\frac{f\left(1\right)-f\left(0\right)}{1}$

$$\begin{gathered} \Rightarrow 2-2x=\frac{1-0}{1} \\ \Rightarrow -2x=1-2 \\ \Rightarrow x=\frac{1}{2} \end{gathered}$$
Thus, $c=\frac{1}{2}\in \left(0,1\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$.

Hence, Lagrange's theorem is verified.

(viii) We have,

$f\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)$ which can be rewritten as $f\left(x\right)=x^3-6x^2+11x-6$

Since a polynomial function is everywhere continuous and differentiable.

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

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

∴ $f\text{'}\left(x\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}$

$$\begin{gathered} \Rightarrow 3x^2-12x+11=\frac{6+6}{4} \\ \Rightarrow 3x^2-12x+8=0 \\ \Rightarrow x=2-\frac{2}{\sqrt{3}},2+\frac{2}{\sqrt{3}} \end{gathered}$$
Thus, $c=2\pm \frac{2}{\sqrt{3}}\in \left(0,4\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}$.

Hence, Lagrange's theorem is verified.

(ix) We have,

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

Here, $f\left(x\right)$ will exist,
if
$$\begin{gathered} 25-x^2\ge 0 \\ \Rightarrow x^2\le 25 \\ \Rightarrow -5\le x\le 5 \end{gathered}$$

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

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

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

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

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

$f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(-3\right)}{4+3}=\frac{f\left(4\right)-f\left(-3\right)}{7}$

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

$f\text{'}\left(x\right)=\frac{-x}{\sqrt{25-x^2}}$ , $f\left(-3\right)=4$ , $f\left(4\right)=3$

∴ $f\text{'}\left(x\right)=\frac{f\left(4\right)-f\left(-3\right)}{4+3}$

$$\begin{gathered} \Rightarrow \frac{-x}{\sqrt{25-x^2}}=\frac{3-4}{7} \\ \Rightarrow 49x^2=25-x^2 \\ \Rightarrow x=\pm \frac{1}{\sqrt{2}} \end{gathered}$$
Thus, $c=\pm \frac{1}{\sqrt{2}}\in \left(-3,4\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(-3\right)}{4-\left(-3\right)}$.

Hence, Lagrange's theorem is verified.

(x) We have,

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

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

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

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

$f\text{'}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}=\frac{f\left(1\right)-f\left(0\right)}{1}$

Now, $f\left(x\right)={\tan }^{-1}x$

$f\text{'}\left(x\right)=\frac{1}{1+x^2}$ , $f\left(1\right)=\frac{\mathrm{\pi }}{4}$ , $f\left(0\right)=0$

∴ $f\text{'}\left(x\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$

$$\begin{gathered} \Rightarrow \frac{1}{1+x^2}=\frac{\mathrm{\pi }}{4}-0 \\ \Rightarrow \left(\frac{4}{\mathrm{\pi }}-1\right)=x^2 \\ \Rightarrow x=\pm \sqrt{\frac{4-\mathrm{\pi }}{\mathrm{\pi }}} \end{gathered}$$
Thus, $c=\sqrt{\frac{4-\mathrm{\pi }}{\mathrm{\pi }}}\in \left(0,1\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$.

Hence, Lagrange's theorem is verified.

(xi) We have,

$f\left(x\right)=x+\frac{1}{x}=\frac{x^2+1}{x}$


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

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

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

$f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}=\frac{f\left(3\right)-f\left(1\right)}{2}$

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

$f\text{'}\left(x\right)=\frac{x^2-1}{x^2}$ , $f\left(1\right)=2$ , $f\left(3\right)=\frac{10}{3}$

∴ $f\text{'}\left(x\right)=\frac{f\left(3\right)-f\left(1\right)}{2}$

$$\begin{gathered} \Rightarrow \frac{x^2-1}{x^2}=\frac{4}{6} \\ \Rightarrow \frac{x^2-1}{x^2}=\frac{2}{3} \\ \Rightarrow 3x^2-3=2x^2 \\ \Rightarrow x=\pm \sqrt{3} \end{gathered}$$
Thus, $c=\sqrt{3}\in \left(1,3\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}$.

Hence, Lagrange's theorem is verified.

(xii) We have,

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

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

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

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

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

$f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}=\frac{f\left(4\right)-f\left(0\right)}{4}$

Now, $f\left(x\right)=x^3+8x^2+16x$

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

∴ $f\text{'}\left(x\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}$

$$\begin{gathered} \Rightarrow 3x^2+16x+16=\frac{256}{4} \\ \Rightarrow 3x^2+16x-48=0 \\ \Rightarrow x=-\frac{4}{3}\left(2+\sqrt{13}\right),\frac{4}{3}\left(\sqrt{13}-2\right) \end{gathered}$$
Thus, $c=\frac{-8+4\sqrt{13}}{3}\in \left(0,4\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}$.

Hence, Lagrange's theorem is verified.

(xiii) We have,

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

Here, $f\left(x\right)$ will exist,
if
$$\begin{gathered} x^2-4\ge 0 \\ \Rightarrow x\le -2orx\ge 2 \end{gathered}$$

Since for each $x\in \left[2,4\right]$, the function $f\left(x\right)$ attains a unique definite value.
So, $f\left(x\right)$ is continuous on $\left[2,4\right]$

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

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

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

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

$f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(2\right)}{4-2}=\frac{f\left(4\right)-f\left(2\right)}{2}$

Now, $f\left(x\right)=\sqrt{x^2-4}$

$f\text{'}\left(x\right)=\frac{x}{\sqrt{x^2-4}}$ , $f\left(4\right)=2\sqrt{3}$ , $f\left(2\right)=0$

∴ $f\text{'}\left(x\right)=\frac{f\left(4\right)-f\left(2\right)}{4-2}$

$$\begin{gathered} \Rightarrow \frac{x}{\sqrt{x^2-4}}=\frac{2\sqrt{3}}{2} \\ \Rightarrow \frac{x}{\sqrt{x^2-4}}=\sqrt{3} \\ \Rightarrow \frac{x^2}{x^2-4}=3 \\ \Rightarrow x^2=3x^2-12 \\ \Rightarrow x^2=6 \\ \Rightarrow x=\pm \sqrt{6} \end{gathered}$$
Thus, $c=\sqrt{6}\in \left(2,4\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(2\right)}{4-2}$.

Hence, Lagrange's theorem is verified.

(xiv) We have,

$f\left(x\right)=x^2+x-1$

Since polynomial function is everywhere continuous and differentiable.

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

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

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

$f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}=\frac{f\left(4\right)-f\left(0\right)}{4}$

Now, $f\left(x\right)=x^2+x-1$

$f\text{'}\left(x\right)=2x+1$ , $f\left(4\right)=19$ , $f\left(0\right)=-1$

∴ $f\text{'}\left(x\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}$

$$\begin{gathered} \Rightarrow 2x+1=\frac{20}{4} \\ \Rightarrow 2x+1=5 \\ \Rightarrow 2x=4 \\ \Rightarrow x=2 \end{gathered}$$
Thus, $c=2\in \left(0,4\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}$.

Hence, Lagrange's theorem is verified.

(xv) We have,

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

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

Therefore, $f\left(x\right)$ is continuous on $\left[0,\mathrm{\pi }\right]$ and differentiable on $\left(0,\mathrm{\pi }\right)$

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

Consequently, there exists some $c\in \left(0,\mathrm{\pi }\right)$ such that

$f\text{'}\left(c\right)=\frac{f\left(\mathrm{\pi }\right)-f\left(0\right)}{\mathrm{\pi }-0}=\frac{f\left(\mathrm{\pi }\right)-f\left(0\right)}{\mathrm{\pi }}$

Now, $f\left(x\right)=\sin x-\sin 2x-x$

$f\text{'}\left(x\right)=\cos x-2\cos 2x-1$ , $f\left(\mathrm{\pi }\right)=-\mathrm{\pi }$ , $f\left(0\right)=0$

∴ $f\text{'}\left(x\right)=\frac{f\left(\mathrm{\pi }\right)-f\left(0\right)}{\mathrm{\pi }-0}$

$$\begin{gathered} \Rightarrow \cos x-2\cos 2x-1=-1 \\ \Rightarrow \cos x-2\cos 2x=0 \\ \Rightarrow \cos x-4{\cos }^{2}x=-2 \\ \Rightarrow 4{\cos }^{2}x-\cos x-2=0 \\ \Rightarrow \cos x=\frac{1}{8}\left(1\pm \sqrt{33}\right) \\ \Rightarrow x={\cos }^{-1}\left[\frac{1}{8}\left(1\pm \sqrt{33}\right)\right] \end{gathered}$$
Thus, $c={\cos }^{-1}\left(\frac{1\pm \sqrt{33}}{8}\right)\in \left(0,\mathrm{\pi }\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(\mathrm{\pi }\right)-f\left(0\right)}{\mathrm{\pi }-0}$.

Hence, Lagrange's theorem is verified.

(xvi) We have,

$f\left(x\right)=x^3-5x^2-3x$

Since polynomial function is everywhere continuous and differentiable

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

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

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

$f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}=\frac{f\left(3\right)-f\left(1\right)}{2}$

Now, $f\left(x\right)=x^3-5x^2-3x$

$f\text{'}\left(x\right)=3x^2-10x-3$ , $f\left(3\right)=-27$ , $f\left(1\right)=-7$

∴ $f\text{'}\left(x\right)=\frac{f\left(3\right)-f\left(1\right)}{2}$
$$\begin{gathered} \Rightarrow 3x^2-10x-3=\frac{-20}{2} \\ \Rightarrow 3x^2-10x+7=0 \\ \Rightarrow x=1,\frac{7}{3} \end{gathered}$$

Thus, $c=\frac{7}{3}\in \left(1,3\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}$.

Hence, Lagrange's theorem is verified.