Chapter 3.3, Problem 9E

To determine

To calculate: The horizontal tangents of the curve.

Answer to Problem 9E

The required horizontal tangents of the curve are $x=0$ and $x=\pm \sqrt{2}$ .

Explanation of Solution

Given information:

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

Formula used:

Power rule:

  $\frac{d}{dx}{x}^n=n{x}^{n-1}$

Quadratic formula:

The roots of the quadratic equation $a{x}^2+bx+c=0$ can be found by

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Calculation:

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

Use the power rule on the above curve.

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(x^4-4x^2+1\right)\\ =\frac{d}{dx}\left(x^4\right)-\frac{d}{dx}\left(4x^2\right)+\frac{d}{dx}\left(1\right)\\ =4x^3-8x+0\\ =4x^3-8x\end{array}$

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

  $4x^3-8x=0$

Factorize the above expression.

  $\begin{array}{c}4x^3-8x=0\\ 4x\left(x^2-2\right)=0\end{array}$

Simplify the above expression for x .

  $\begin{array}{c}4x=0\\ x=0\end{array}$

And,

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

Thus, the roots of the expression are $x=0$ and $x=\pm \sqrt{2}$ .

Hence, the required horizontal tangents of the curve are $x=0$ and $x=\pm \sqrt{2}$ .