Question

The equation of tangent to the curve $y = x + \dfrac{4}{{{x^2}}}$, that is parallel to $x - {\text{axis}}$ is,
A) $y = 1$
B) $y = 2$
C) $y = 3$
D) $y = 0$

Answer 1

The slope of equation of the tangent to the curve $y = f\left( x \right)$ is given by $\dfrac{{dy}}{{dx}}$ that means differentiating with respect to $x$. If the equation of tangent is parallel to $x$, then its slope must be $0$.

Complete step-by-step answer:
Here we are given an equation of curve, that is $y = x + \dfrac{4}{{{x^2}}}$, and condition is given that it must be parallel to $x - {\text{axis}}$. So if any line is parallel to $x - {\text{axis}}$, then its slope must be $0$.
So, we are given
$y = x + \dfrac{4}{{{x^2}}}$
Now, differentiating with respect to $x$, we get
$\dfrac{{dy}}{{dx}} = \dfrac{{dx}}{{dx}} + 4\dfrac{{d\left( {\dfrac{1}{{{x^2}}}} \right)}}{{dx}} \\\ \dfrac{{dy}}{{dx}} = 1 + 4\left( { - 2} \right){x^{ - 2 - 1}} \\\ \dfrac{{dy}}{{dx}} = 1 - 8{x^{ - 3}} \\\ \dfrac{{dy}}{{dx}} = 1 - \dfrac{8}{{{x^3}}} \\\$
Now, we know that $\dfrac{{dy}}{{dx}}$ is the slope of the tangent to that curve. And as it is saying that the equation of tangent when it is parallel to $x - {\text{axis}}$, so its slope must be zero.
So,
$\dfrac{{dy}}{{dx}} = 0 \\\ 1 - \dfrac{8}{{{x^3}}} = 0 \\\ \dfrac{8}{{{x^3}}} = 1 \\\ {x^3} = 8 \\\ x = {8^{\dfrac{1}{3}}} = {\left( {{2^3}} \right)^{\dfrac{1}{3}}} \\\ x = 2 \\\$
So, we know that $y = x + \dfrac{4}{{{x^2}}}$
So,
$y = 2 + \dfrac{4}{{{2^2}}} \\\ = 2 + \dfrac{4}{4} = 2 + 1 \\\ y = 3 \\\$
So, $\left( {2,3} \right)$ is the point of contact. So, the equation of tangent will be $y = 3$.

So, the correct answer is “Option C”.

Note: We know that if we need to differentiate $\dfrac{1}{{{x^n}}}$ with respect to $x$, then we get $- n{x^{ - n - 1}}$.If we differentiate any curve or any derivative of function, we get slope of equation of the tangent.Here in this question the equation of tangent which is parallel to x axis means the slope of equation of tangent is 0 i.e. $\dfrac{{dy}}{{dx}} = 0$.