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# How do you find the points where the tangent line is horizontal given $y = 16{x^{ - 1}} - {x^2}$?

Last updated date: 07th Aug 2024
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Hint: In the given question, we have been given an equation of a tangent line. We have to find the points where this line is horizontal. We are going to solve it by first simplifying the equation. Then we are going to find the derivative of the simplified equation of the given line. Then we are going to solve the equation. Then we are going to put the calculated points back into the original equation and find the value of the other variable. And that is going to give us our answer.

Formula Used:
We are going to use the formula of derivative, which is,
$\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$

Complete step by step solution:
The given equation is $y = 16{x^{ - 1}} - {x^2}$.
First, we are going to find the derivative of the given equation,
$\dfrac{{dy}}{{dx}} = \dfrac{{d\left( {16{x^{ - 1}} - {x^2}} \right)}}{{dx}}$
We are going to use the formula of derivative, which is,
$\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$
So, we have,
$\dfrac{{dy}}{{dx}} = - 16{x^{ - 2}} - 2x$
Now, we are going to put it equal to zero and calculate the points,
$- \dfrac{{16}}{{{x^2}}} - 2x = 0$
So, we have,
$- \dfrac{{16}}{{{x^2}}} = 2x$
$\Rightarrow {x^3} = - 8$
Hence, $x = - 2$
Now, let us put the calculated value back into the original equation,
$y = \dfrac{{16}}{{\left( { - 2} \right)}} - {\left( { - 2} \right)^2} = - 8 - 4 = - 12$

Thus, the required point is $\left( { - 2, - 12} \right)$.

Note:
In the given question, we had to find the points where the given tangent line was horizontal. We solved it by first finding the derivative of the given line. Then we computed the value of the derivative by putting it equal to zero and found the value of the variable. Then we put the value of the variable back into the original equation and found the points.