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# Verify Lagrange’s mean value theorem for the following function on the indicated interval. In each case find a point $'c'$ in the indicated interval as stated by Lagrange’s mean value theorem: $f\left( x \right) = {x^3} - 5x - 3$ on $\left[ {1,3} \right]$  Verified
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Hint: So for solving this type of question we just need to apply Lagrange’s mean value theorem condition which is $f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$ for a function and then we need to simplify the function so that we will calculate the derivative, and it will be the required theorem which we have verified.
Formula used:
Lagrange’s mean value theorem given by,
$f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$
Such that $c$ will lie in the interval of $\left[ {a,b} \right]$ .

So we have the function given as $f\left( x \right) = {x^3} - 5x - 3$ .
Since we know that the polynomial function is everywhere continuous and differentiable. Therefore, $f\left( x \right)$ will be continuous on the interval $\left[ {1,3} \right]$ and also will be differentiable on $\left( {1,3} \right)$ . And in both conditions, Lagrange’s mean value theorem will be satisfied.
So by using Lagrange’s mean value theorem,
$\Rightarrow f'\left( c \right) = \dfrac{{f\left( 3 \right) - f\left( 1 \right)}}{{3 - 1}}$
And on solving the denominator we get
$\Rightarrow f'\left( c \right) = \dfrac{{f\left( 3 \right) - f\left( 1 \right)}}{2}$
Now we have $f\left( x \right) = {x^3} - 5x - 3$
On differentiating it with respect to $x$ , we get the equation as
$\Rightarrow f'\left( x \right) = 3{x^2} - 5$
Now on solving for the value of $x = 1$ in $f\left( x \right) = {x^3} - 5x - 3$ , we get
$\Rightarrow f\left( 1 \right) = - 7$
Now on solving for the value of $x = 3$ in $f\left( x \right) = {x^3} - 5x - 3$ , we get
$\Rightarrow f\left( 3 \right) = 9$
Therefore, $f'\left( x \right) = \dfrac{{f\left( 3 \right) - f\left( 1 \right)}}{2}$
On substituting the values, we get
$\Rightarrow 3{x^2} - 5 = 8$
Now taking the constant term to the right side, we get
$\Rightarrow 3{x^2} = 8 - 5$
And on solving it, we get
$\Rightarrow {x^2} = 1$
And therefore, $x = 1$ which comes in the interval such that $f'\left( c \right) = \dfrac{{f\left( 3 \right) - f\left( 1 \right)}}{2}$
Hence, Lagrange’s theorem is verified.

Note: Here, in this type of question there are very few points where we can make mistakes like not applying the differentiation rules for the derivation of the mentioned function. Since in this question, the equation is simple but if we have the complex equation then we should use the determinant formula which is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ .