Question

Find the points where the gradient of the curve \[y = 12x - {x^3}\] is 0.
A. (0,2),(2,16)
B. (0,-2),(2,-16)
C. (2,-16),(-2,16)
D. (2,16),(-2,-16)

Answer 1

Hint First we will find the derivative of \[y = 12x - {x^3}\] with respect to \[x\]. Then we equate \[\dfrac{{dy}}{{dx}}\] with zero. From the equation \[\dfrac{{dy}}{{dx}} = 0\], we calculate the value of \[x\] and put the value \[x\] in \[y = 12x - {x^3}\] to get the value of \[y\]. The solutions of \[\left( {x,y} \right)\] are required points.

Formula Used
\[\dfrac{d}{{dx}}\left( {g\left( x \right) \pm f\left( x \right)} \right) = \dfrac{d}{{dx}}g\left( x \right) \pm \dfrac{d}{{dx}}f\left( x \right)\]
\[\dfrac{d}{{dx}}\left[ {kf\left( x \right)} \right] = k\dfrac{d}{{dx}}f\left( x \right)\] where \[k\] is a constant.
\[\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}\]

Complete step by step solution
Given equation of the curve is \[y = 12x - {x^3}\].
Differentiate the curve with respect to \[x\].
\[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {12x - {x^3}} \right)\]
Apply the formula \[\dfrac{d}{{dx}}\left( {g\left( x \right) \pm f\left( x \right)} \right) = \dfrac{d}{{dx}}g\left( x \right) \pm \dfrac{d}{{dx}}f\left( x \right)\]
\[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {12x} \right) - \dfrac{d}{{dx}}\left( {{x^3}} \right)\]
Apply the formula \[\dfrac{d}{{dx}}\left[ {kf\left( x \right)} \right] = k\dfrac{d}{{dx}}f\left( x \right)\]
\[\dfrac{{dy}}{{dx}} = 12\dfrac{d}{{dx}}\left( x \right) - \dfrac{d}{{dx}}\left( {{x^3}} \right)\]
Apply the formula \[\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}\]
\[\dfrac{{dy}}{{dx}} = 12 \cdot 1 - 3{x^2}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = 12 - 3{x^2}\]
Now we will equate \[\dfrac{{dy}}{{dx}}\] with aero.
\[12 - 3{x^2} = 0\]
Solve the above equation
\[ \Rightarrow 12 = 3{x^2}\]
Divide both sides by \[3\].
\[ \Rightarrow 4 = {x^2}\]
Taking square root both sides
\[x = \pm 2\]
Now we will put \[x = 2\] in the equation \[y = 12x - {x^3}\]
\[y = 12 \cdot 2 - {2^3}\]
\[ \Rightarrow y = 24 - 8\]
\[ \Rightarrow y = 16\]
Now we will put \[x = - 2\] in the equation \[y = 12x - {x^3}\]
\[y = 12 \cdot \left( { - 2} \right) - {\left( { - 2} \right)^3}\]
\[ \Rightarrow y = - 24 + 8\]
\[ \Rightarrow y = - 16\]
The points are (2,16),(-2,-16).
Hence the correct option is option C.

Note Students often make a common mistake that they solve the equation \[{x^2} = 4\] as \[x = 2\]. They do not consider the negative value. There are two roots of a quadratic equation. Whenever we apply a square root to an equation, we should use a \[ \pm \] sign.