Question

Find intervals in which $f\left( x \right) = \dfrac{{4{x^2} + 1}}{x}$ is increasing or decreasing.

Answer 1

Here, we need to find the intervals in which the given function is increasing or decreasing. We will use differentiation to find the critical points where the first derivative is equal to 0. Then, we will divide the set of all real numbers at the critical points. Finally, we will check the sign of the first derivative at any value from the intervals to find where the function is increasing or decreasing.

Formula Used:
We will use the following formulas:
1.The derivative of the sum of two functions is the sum of the derivatives of the two functions, that is $\dfrac{{d\left[ {f\left( x \right) + g\left( x \right)} \right]}}{{dx}} = \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} + \dfrac{{d\left[ {g\left( x \right)} \right]}}{{dx}}$.
2.The derivative of a function of the form $\dfrac{{d\left[ {af\left( x \right)} \right]}}{{dx}}$ can be written as $a\dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}}$.
3.The derivative of a function of the form ${x^n}$ is $n{x^{n - 1}}$.

Complete step-by-step answer:
First, we will differentiate the given function using the quotient rule of differentiation.
Differentiating both sides of $f\left( x \right) = \dfrac{{4{x^2} + 1}}{x}$ with respect to $x$ using the quotient rule, we get
$\Rightarrow f'\left( x \right) = \dfrac{{x\dfrac{{d\left( {4{x^2} + 1} \right)}}{{dx}} - \left( {4{x^2} + 1} \right)\dfrac{{d\left( x \right)}}{{dx}}}}{{{x^2}}}$
The derivative of the sum of two functions is the sum of the derivatives of the two functions, that is $\dfrac{{d\left[ {f\left( x \right) + g\left( x \right)} \right]}}{{dx}} = \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} + \dfrac{{d\left[ {g\left( x \right)} \right]}}{{dx}}$.
Therefore, we can rewrite $\dfrac{{d\left( {4{x^2} + 1} \right)}}{{dx}}$ as $\dfrac{{d\left( {4{x^2}} \right)}}{{dx}} + \dfrac{{d\left( 1 \right)}}{{dx}}$.
Thus, the equation becomes
$\Rightarrow f'\left( x \right) = \dfrac{{x\left[ {\dfrac{{d\left( {4{x^2}} \right)}}{{dx}} + \dfrac{{d\left( 1 \right)}}{{dx}}} \right] - \left( {4{x^2} + 1} \right)\dfrac{{d\left( x \right)}}{{dx}}}}{{{x^2}}}$
The derivative of a function of the form $\dfrac{{d\left[ {af\left( x \right)} \right]}}{{dx}}$ can be written as $a\dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}}$.
Therefore, we can rewrite $\dfrac{{d\left( {4{x^2}} \right)}}{{dx}}$ as $4\dfrac{{d\left( {{x^2}} \right)}}{{dx}}$.
Thus, we get
$\Rightarrow f'\left( x \right) = \dfrac{{x\left[ {4\dfrac{{d\left( {{x^2}} \right)}}{{dx}} + \dfrac{{d\left( 1 \right)}}{{dx}}} \right] - \left( {4{x^2} + 1} \right)\dfrac{{d\left( x \right)}}{{dx}}}}{{{x^2}}}$
The derivative of a function of the form ${x^n}$ is $n{x^{n - 1}}$.
Therefore, we get
$\begin{array}{l} \Rightarrow \dfrac{{d\left( {{x^2}} \right)}}{{dx}} = 2{x^{2 - 1}}\\\ \Rightarrow \dfrac{{d\left( {{x^2}} \right)}}{{dx}} = 2{x^1}\\\ \Rightarrow \dfrac{{d\left( {{x^2}} \right)}}{{dx}} = 2x\end{array}$
The derivative of a constant is always 0.
Therefore, we get
$\Rightarrow \dfrac{{d\left( 1 \right)}}{{dx}} = 0$
The derivative of a variable with respect to itself is always 1.
Therefore, we get
$\Rightarrow \dfrac{{d\left( x \right)}}{{dx}} = 1$
Substituting $\dfrac{{d\left( {{x^2}} \right)}}{{dx}} = 2x$, $\dfrac{{d\left( 1 \right)}}{{dx}} = 0$, and $\dfrac{{d\left( x \right)}}{{dx}} = 1$ in the equation $f'\left( x \right) = \dfrac{{x\left[ {4\dfrac{{d\left( {{x^2}} \right)}}{{dx}} + \dfrac{{d\left( 1 \right)}}{{dx}}} \right] - \left( {4{x^2} + 1} \right)\dfrac{{d\left( x \right)}}{{dx}}}}{{{x^2}}}$, we get
$\Rightarrow f'\left( x \right) = \dfrac{{x\left[ {4\left( {2x} \right) + 0} \right] - \left( {4{x^2} + 1} \right)\left( 1 \right)}}{{{x^2}}}$
Multiplying the terms in the expression, we get
$\begin{array}{l} \Rightarrow f'\left( x \right) = \dfrac{{x\left[ {8x} \right] - \left( {4{x^2} + 1} \right)}}{{{x^2}}}\\\ \Rightarrow f'\left( x \right) = \dfrac{{8{x^2} - \left( {4{x^2} + 1} \right)}}{{{x^2}}}\end{array}$
Simplifying the numerator of the expression, we get
$\begin{array}{l} \Rightarrow f'\left( x \right) = \dfrac{{8{x^2} - 4{x^2} - 1}}{{{x^2}}}\\\ \Rightarrow f'\left( x \right) = \dfrac{{4{x^2} - 1}}{{{x^2}}}\end{array}$
Now, we will find the critical points of the function.
The critical points are the points at which $f'\left( x \right) = 0$.
Substituting $f'\left( x \right) = \dfrac{{4{x^2} - 1}}{{{x^2}}}$ in $f'\left( x \right) = 0$, we get
$\Rightarrow \dfrac{{4{x^2} - 1}}{{{x^2}}} = 0$
Multiplying both sides of the equation by ${x^2}$, we get
$\Rightarrow 4{x^2} - 1 = 0$
Rewriting the expression, we get
$\Rightarrow {\left( {2x} \right)^2} - {1^2} = 0$
Using the algebraic identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$, we get
$\Rightarrow \left( {2x + 1} \right)\left( {2x - 1} \right) = 0$
Therefore, we get
$\Rightarrow 2x - 1 = 0$ or $2x - 1 = 0$
Simplifying the expressions, we get
$\Rightarrow x = - \dfrac{1}{2},\dfrac{1}{2}$
Therefore, we get the critical points as $x = - \dfrac{1}{2},\dfrac{1}{2}$.
We will divide the set of all real numbers $\left( { - \infty ,\infty } \right)$ at the critical points.
Therefore, we get the intervals $\left( { - \infty , - \dfrac{1}{2}} \right)$, $\left( { - \dfrac{1}{2},\dfrac{1}{2}} \right)$, and $\left( {\dfrac{1}{2},\infty } \right)$.
A function $f\left( x \right)$ is increasing at an interval if $f'\left( x \right) > 0$ for all values of $x$ belonging to that interval.
A function $f\left( x \right)$ is decreasing at an interval if $f'\left( x \right) < 0$ for all values of $x$ belonging to that interval.
We will check the sign of $f'\left( x \right) = \dfrac{{4{x^2} - 1}}{{{x^2}}}$ at some point within the intervals $\left( { - \infty , - \dfrac{1}{2}} \right)$, $\left( { - \dfrac{1}{2},\dfrac{1}{2}} \right)$, and $\left( {\dfrac{1}{2},\infty } \right)$.
(i) We will take a value from the interval $\left( { - \infty , - \dfrac{1}{2}} \right)$.
Let $x = - 1$.
Substituting $x = - 1$ in the equation $f'\left( x \right) = \dfrac{{4{x^2} - 1}}{{{x^2}}}$, we get
$\Rightarrow f'\left( x \right) = \dfrac{{4{{\left( { - 1} \right)}^2} - 1}}{{{{\left( { - 1} \right)}^2}}}$
Simplifying the expression, we get
$\begin{array}{l} \Rightarrow f'\left( x \right) = \dfrac{{4\left( 1 \right) - 1}}{1}\\\ \Rightarrow f'\left( x \right) = 4 - 1\\\ \Rightarrow f'\left( x \right) = 3\end{array}$
Therefore, we can see that if $x \in \left( { - \infty , - \dfrac{1}{2}} \right)$, then $f'\left( x \right) > 0$.
Thus, the function $f\left( x \right) = \dfrac{{4{x^2} + 1}}{x}$ is increasing in the interval $\left( { - \infty , - \dfrac{1}{2}} \right)$.
(ii) We will take a value from the interval $\left( { - \dfrac{1}{2},\dfrac{1}{2}} \right)$.
Let $x = \dfrac{1}{4}$.
Substituting $x = \dfrac{1}{4}$ in the equation $f'\left( x \right) = \dfrac{{4{x^2} - 1}}{{{x^2}}}$, we get
$\Rightarrow f'\left( x \right) = \dfrac{{4{{\left( {\dfrac{1}{4}} \right)}^2} - 1}}{{{{\left( {\dfrac{1}{4}} \right)}^2}}}$
Multiplying the terms, we get
$\begin{array}{l} \Rightarrow f'\left( x \right) = \dfrac{{4\left( {\dfrac{1}{{16}}} \right) - 1}}{{\left( {\dfrac{1}{{16}}} \right)}}\\\ \Rightarrow f'\left( x \right) = \dfrac{{\dfrac{4}{{16}} - 1}}{{\dfrac{1}{{16}}}}\end{array}$
Simplifying the expression, we get
$\begin{array}{l} \Rightarrow f'\left( x \right) = \dfrac{{\dfrac{{ - 12}}{{16}}}}{{\dfrac{1}{{16}}}}\\\ \Rightarrow f'\left( x \right) = - 12\end{array}$
Therefore, we can see that if $x \in \left( { - \dfrac{1}{2},\dfrac{1}{2}} \right)$, then $f'\left( x \right) < 0$.
Thus, the function $f\left( x \right) = \dfrac{{4{x^2} + 1}}{x}$ is decreasing in the interval $\left( { - \dfrac{1}{2},\dfrac{1}{2}} \right)$.
(iii) We will take a value from the interval $\left( {\dfrac{1}{2},\infty } \right)$.
Let $x = 1$.
Substituting $x = 1$ in the equation $f'\left( x \right) = \dfrac{{4{x^2} - 1}}{{{x^2}}}$, we get
$\Rightarrow f'\left( x \right) = \dfrac{{4{{\left( 1 \right)}^2} - 1}}{{{{\left( 1 \right)}^2}}}$
Simplifying the expression, we get
$\begin{array}{l} \Rightarrow f'\left( x \right) = \dfrac{{4\left( 1 \right) - 1}}{1}\\\ \Rightarrow f'\left( x \right) = 4 - 1\\\ \Rightarrow f'\left( x \right) = 3\end{array}$
Therefore, we can see that if $x \in \left( {\dfrac{1}{2},\infty } \right)$, then $f'\left( x \right) > 0$.
Thus, the function $f\left( x \right) = \dfrac{{4{x^2} + 1}}{x}$ is increasing in the interval $\left( {\dfrac{1}{2},\infty } \right)$.
Hence, we get that the function $f\left( x \right) = \dfrac{{4{x^2} + 1}}{x}$ is increasing in the intervals $\left( { - \infty , - \dfrac{1}{2}} \right)$ and $\left( {\dfrac{1}{2},\infty } \right)$, and decreasing in the interval $\left( { - \dfrac{1}{2},\dfrac{1}{2}} \right)$.

Note: We used the quotient rule of differentiation in the solution. The quotient rule states that the derivative of a function of the form $y=\dfrac{u}{v}$ with respect to $x$ is given by the formula $\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{v\dfrac{{d\left( u \right)}}{{dx}} - u\dfrac{{d\left( v \right)}}{{dx}}}}{{{v^2}}}$.
We used the algebraic identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ in the solution. The difference of the squares of two numbers is given by the algebraic identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$.