Question

The abscissa of the points, where the tangent to the curve $y={{x}^{3}}-3{{x}^{2}}-9x+5$ is parallel to the x – axis, are?
(a) $x$ = 0 and 0
(b) $x$ = 1 and -1
(c) $x$ = 1 and -3
(d) $x$ = -1 and 3

Answer 1

Hint: Find the derivative of the curve by differentiating the function both the sides with respect to x and using the formula $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$. Consider the slope of the line parallel to the x – axis equal to 0 and substitute $\dfrac{dy}{dx}=0$. Solve the quadratic equation by splitting the middle term method and find the values of x to get the answer.

Complete step by step solution:
Here we have been provided with the curve $y={{x}^{3}}-3{{x}^{2}}-9x+5$ and we are asked to find the abscissa (x – coordinate) of the point where the tangent to the curve is parallel to the x – axis.
Now, in mathematics the slope of tangent $\left( \dfrac{dy}{dx} \right)$ of a curve at a point is the value of the derivative of the curve at that point. So differentiating the given function both the sides with respect to x we get,
$\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( {{x}^{3}}-3{{x}^{2}}-9x+5 \right)}{dx} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( {{x}^{3}} \right)}{dx}-\dfrac{d\left( 3{{x}^{2}} \right)}{dx}-\dfrac{d\left( 9x \right)}{dx}+\dfrac{d\left( 5 \right)}{dx} \\
\end{align}$
We know that the derivative of a constant is 0. When a constant is multiplied with a variable then the constant term can be taken out of the derivative. Also, using the formula $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$ we get,
$\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=3{{x}^{2}}-6x-9+0 \\
 & \Rightarrow \dfrac{dy}{dx}=3{{x}^{2}}-6x-9 \\
\end{align}$
Now, any line parallel to the x – axis has a slope equal to 0, so equating $\dfrac{dy}{dx}=0$ we get,
$\begin{align}
  & \Rightarrow 3{{x}^{2}}-6x-9=0 \\
 & \Rightarrow {{x}^{2}}-2x-3=0 \\
\end{align}$
Applying the middle term split method to factor the terms of the above quadratic equation we get,
$\begin{align}
  & \Rightarrow {{x}^{2}}-3x+x-3=0 \\
 & \Rightarrow \left( x+1 \right)\left( x-3 \right)=0 \\
\end{align}$
Substituting each term equal to 0 we get,
$\Rightarrow \left( x+1 \right)=0$ or $\left( x-3 \right)=0$
$\Rightarrow x=-1$ or $x=3$
Hence, option (d) is the correct answer.

Note: : Always remember that the slope of a line parallel to the x – axis is equal to 0 and that of a line parallel to the y – axis is infinite. If two lines are perpendicular to each other then the product of their slopes is equal to -1. Remember the formulas of derivatives of functions like: - trigonometric, inverse trigonometric, logarithmic, exponential functions etc.