Chapter 3.3, Problem 10E

To determine

To calculate: The horizontal tangents of the curve.

Answer to Problem 10E

The required horizontal tangents of the curve are $x=0$ and $x=1$ .

Explanation of Solution

Given information:

The curve: $y=4x^3-6x^2-1$

Formula used:

Power rule:

  $\frac{d}{dx}{x}^n=n{x}^{n-1}$

Calculation:

The given curve is $y=4x^3-6x^2-1$ .

Use the power rule on the above curve.

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(4x^3-6x^2-1\right)\\ =\frac{d}{dx}\left(4x^3\right)-\frac{d}{dx}\left(6x^2\right)-\frac{d}{dx}\left(1\right)\\ =12x^2-12x-0\\ =12x^2-12x\end{array}$

Now, to find the horizontal tangents of the given curve, set the above derivative equal to zero.

  $12x^2-12x=0$

Factorize the above expression and solve for the roots of the expression.

  $\begin{array}{c}12x^2-12x=0\\ \frac{\cancel{12}{x}^2}{\cancel{12}}-\frac{\cancel{12}x}{\cancel{12}}=0\\ x^2-x=0\\ x\left(x-1\right)=0\end{array}$

Simplify the above expression further.

  $x=0$

And

  $\begin{array}{c}x-1=0\\ x=1\end{array}$

Thus, the roots of the expression are $x=0,1$ .

Hence, the required horizontal tangents of the curve are $x=0$ and $x=1$ .