Question

# If $f(x) = {x^2} - x + 5$, $x > \dfrac{1}{2}$ and $g(x)$ is its inverse function, then $g'(7)$ equalsA. $\dfrac{{ - 1}}{3}$B. $\dfrac{1}{{13}}$C. $\dfrac{1}{3}$D. $\dfrac{{ - 1}}{{13}}$

Hint: Here we will use the concept that inverse of a function gives the value on which we are applying the function, i.e. ${f^{ - 1}}(f(x)) = x$ and then find the differentiation of g with respect to x. Substituting the value of $f(x)$ as 7 we calculate the value of x and substitute in the differentiation of g.
* Chain rule of differentiation is given as $\dfrac{d}{{dx}}f(g(x)) = f'(g(x)) \times g'(x)$ where $f'(g(x))$ is differentiation of f and $g'(x)$ is differentiation of g with respect to x.

We have the function $f(x) = {x^2} - x + 5$
We find the differentiation of $f(x)$
$\Rightarrow \dfrac{d}{{dx}}f(x) = \dfrac{d}{{dx}}({x^2} - x + 5)$
$\Rightarrow \dfrac{d}{{dx}}f(x) = (2x - 1)$ … (1)
Now we know that $g(x)$ is an inverse function of $f(x)$.
Then we can write that when g is applied to f we get the value x
$\Rightarrow g(f(x)) = x$
Now we differentiate both sides of the equation with respect to x
$\Rightarrow \dfrac{d}{{dx}}g(f(x)) = \dfrac{d}{{dx}}(x)$
Using chain rule on LHS of the equation and differentiating RHS of the equation.
$\Rightarrow g'(f(x)).f'(x) = 1$
Dividing both sides of the equation by $f'(x)$
$\Rightarrow \dfrac{{g'(f(x)).f'(x)}}{{f'(x)}} = \dfrac{1}{{f'(x)}}$
Cancel the same terms from numerator and denominator.
$\Rightarrow g'(f(x)) = \dfrac{1}{{f'(x)}}$
Substituting the value from equation (1), we get
$\Rightarrow g'(f(x)) = \dfrac{1}{{(2x - 1)}}$ … (2)
We know that $g(f(x)) = x$, We have to find the value of $g'(7)$. We find the value of $g(7)$, i.e.
$\Rightarrow f(x) = 7$
Substitute the value of $f(x) = {x^2} - x + 5$
$\Rightarrow {x^2} - x + 5 = 7$
Shift all values to one side of the equation
$\Rightarrow {x^2} - x + 5 - 7 = 0 \\ \Rightarrow {x^2} - x - 2 = 0 \\$
Now we factorize the equation by writing $- x = - 2x + x$
$\Rightarrow {x^2} - 2x + x - 2 = 0$
Take x common from first two terms and 1 common from last two terms
$\Rightarrow x(x - 2) + 1(x - 2) = 0$
Pair up the factors
$\Rightarrow (x - 2)(x + 1) = 0$
Now we can equate both the factors equal to zero.
Firstly $x - 2 = 0$
Shift the constant to one side of the equation.
$\Rightarrow x = 2$
Secondly, $x + 1 = 0$
Shift the constant to one side of the equation.
$\Rightarrow x = - 1$
So, $x = 2,x = - 1$
But we have a condition $x > \dfrac{1}{2}$ i.e. $x > 0.5$
So we have to reject the value of $x = - 1$.
So, our value of $x = 2$.
Now we substitute the value in equation (2)
$\Rightarrow g'(f(2)) = \dfrac{1}{{(2 \times 2 - 1)}} \\ \Rightarrow g'(f(2)) = \dfrac{1}{{(4 - 1)}} \\ \Rightarrow g'(f(2)) = \dfrac{1}{3} \\$
So, the value of $g'(7) = \dfrac{1}{3}$

So, option C is correct.

Note: Students should always keep in mind that the sign changes from negative to positive and vice versa when shifting any value from one side of the equation to another. Also, while applying chain rule of differentiation, always keep in mind that we differentiate the outer function first without looking at the function in the bracket then we differentiate the function inside the bracket.
Students many times make mistake of writing inverse of function by taking the reciprocal of the function, i.e. if $f(x) = 2x + 5$, then they write ${f^{ - 1}}(x) = \dfrac{1}{{2x + 5}}$ which is wrong.