Question

How do you prove from the definition of differentiability that the function $f\left( x \right) = \dfrac{{2x + 1}}{{x - 2}}$ is differentiable?

Answer 1

Hint: This question is based on the definition of differentiability. The definition is: let $f:\mathbb{R} \to \mathbb{R}$, be a function f is differentiable at x, if the following limit exists: $\mathop {\lim }\limits_{t \to x} \dfrac{{f\left( t \right) - f\left( x \right)}}{{t - x}}$, where, $t \in \mathbb{R}$ and $t \ne x$. This equation is called the derivative of f at x, and it is denoted by $f'\left( x \right)$.

Complete step-by-step solution:
In this question, the function is given as below.
$ \Rightarrow f\left( x \right) = \dfrac{{2x + 1}}{{x - 2}}$
Here, $x \in \mathbb{R} - \left\{ 2 \right\}$
Now, the definition of differentiability is as stated below.
 Let $f:\mathbb{R} \to \mathbb{R}$, be a function f is differentiable at x, if the following limit exists: $\mathop {\lim }\limits_{t \to x} \dfrac{{f\left( t \right) - f\left( x \right)}}{{t - x}}$, where, $t \in \mathbb{R}$ and $t \ne x$. This equation is called the derivative of f at x, and it is denoted by $f'\left( x \right)$.
Therefore,
$ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{t \to x} \dfrac{{f\left( t \right) - f\left( x \right)}}{{t - x}}$ ...(1)
Let us know the value of $f\left( t \right)$.
$ \Rightarrow f\left( t \right) = \dfrac{{2t + 1}}{{t - 2}}$
  Where, $t \in \mathbb{R} - \left\{ 2 \right\},t \ne x$
Now, let us find the value of,
$ \Rightarrow f\left( t \right) - f\left( x \right)$
Substitute the values of $f\left( t \right)$ and $f\left( x \right)$.
$ \Rightarrow f\left( t \right) - f\left( x \right) = \dfrac{{2t + 1}}{{t - 2}} - \dfrac{{2x + 1}}{{x - 2}}$
Let us take LCM on the right-hand side.
$ \Rightarrow f\left( t \right) - f\left( x \right) = \dfrac{{\left( {2t + 1} \right)\left( {x - 2} \right) - \left( {2x + 1} \right)\left( {t - 2} \right)}}{{\left( {t - 2} \right)\left( {x - 2} \right)}}$
Let us multiply the numerator to remove the brackets.
$ \Rightarrow f\left( t \right) - f\left( x \right) = \dfrac{{\left( {2tx - 4t + x - 2} \right) - \left( {2tx - 4x + t - 2} \right)}}{{\left( {t - 2} \right)\left( {x - 2} \right)}}$
Let us open the brackets of the numerator.
$ \Rightarrow f\left( t \right) - f\left( x \right) = \dfrac{{2tx - 4t + x - 2 - 2tx + 4x - t + 2}}{{\left( {t - 2} \right)\left( {x - 2} \right)}}$
Let us simplify the above step.
$ \Rightarrow f\left( t \right) - f\left( x \right) = \dfrac{{ - 5t + 5x}}{{\left( {t - 2} \right)\left( {x - 2} \right)}}$
Let us take out -5 as a common factor from the numerator.
$ \Rightarrow f\left( t \right) - f\left( x \right) = \dfrac{{ - 5\left( {t - x} \right)}}{{\left( {t - 2} \right)\left( {x - 2} \right)}}$
Now, let us substitute this value in the equation (1).
$ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{t \to x} \dfrac{{f\left( t \right) - f\left( x \right)}}{{t - x}}$
Put $f\left( t \right) - f\left( x \right) = \dfrac{{ - 5\left( {t - x} \right)}}{{\left( {t - 2} \right)\left( {x - 2} \right)}}$ in the above equation.
$ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{t \to x} \dfrac{{\dfrac{{ - 5\left( {t - x} \right)}}{{\left( {t - 2} \right)\left( {x - 2} \right)}}}}{{t - x}}$
That is equal to,
$ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{t \to x} \dfrac{{ - 5}}{{\left( {t - 2} \right)\left( {x - 2} \right)}}$
Now, let us apply the limit $t \to x$.
$ \Rightarrow f'\left( x \right) = \dfrac{{ - 5}}{{\left( {x - 2} \right)\left( {x - 2} \right)}}$
Let us simplify the denominator on the right-hand side.
$ \Rightarrow f'\left( x \right) = \dfrac{{ - 5}}{{{{\left( {x - 2} \right)}^2}}}$

Hence, we find that the limit in the given function exists. The given function is differentiable at $x \in \mathbb{R} - \left\{ 2 \right\}$, and $ \Rightarrow f'\left( x \right) = \dfrac{{ - 5}}{{{{\left( {x - 2} \right)}^2}}}$, $x \in \mathbb{R} - \left\{ 2 \right\}$.

Note: The function $f\left( x \right)$ is said to be non-differentiable if,
Both right-hand derivative (RHD) and left-hand derivative (LHD) exist but not equal.
Either or both RHD and LHD are not finite.
Either or both RHD and LHD do not exist.