Chapter 3.3, Problem 11E

To determine

To calculate: The horizontal tangents of the curve.

Answer to Problem 11E

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

Explanation of Solution

Given information:

The curve: $y=5x^3-3x^5$

Formula used:

Power rule:

  $\frac{d}{dx}{x}^n=n{x}^{n-1}$

Calculation:

The given curve is $y=5x^3-3x^5$ .

Use the power rule on the above curve.

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(5x^3-3x^5\right)\\ =\frac{d}{dx}\left(5x^3\right)-\frac{d}{dx}\left(3x^5\right)\\ =15x^2-15x^4\end{array}$

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

  $15x^2-15x^4=0$

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

  $\begin{array}{c}15x^2-15x^4=0\\ 15x^2\left(-x^2+1\right)=0\\ \frac{\cancel{15}{x}^2}{\cancel{15}}\left(-x^2+1\right)=0\\ x^2\left(-x^2+1\right)=0\end{array}$

Simplify the above expression further.

  $\begin{array}{c}{x}^2=0\\ x=0\end{array}$

And

  $\begin{array}{c}-x^2+1=0\\ x^2=1\\ \sqrt{x^2}=\pm \sqrt{1}\\ x=\pm 1\end{array}$

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

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