Chapter 3.3, Problem 12E

To determine

To calculate: The horizontal tangents of the curve.

Answer to Problem 12E

The required horizontal tangents of the curve are $x=0$ , $x=5.052$ and $x=0.197$ .

Explanation of Solution

Given information:

The curve: $y=x^4-7x^3+2x^2+15$

Formula used:

Power rule:

  $\frac{d}{dx}{x}^n=n{x}^{n-1}$

Quadratic formula:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

The given curve is $y=x^4-7x^3+2x^2+15$ .

Use the power rule on the above curve.

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(x^4-7x^3+2x^2+15\right)\\ =\frac{d}{dx}\left(x^4\right)-\frac{d}{dx}\left(7x^3\right)+\frac{d}{dx}\left(2x^2\right)+\frac{d}{dx}\left(15\right)\\ =4x^3-21x^2+4x+0\\ =4x^3-21x^2+4x\end{array}$

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

  $4x^3-21x^2+4x=0$

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

  $\begin{array}{c}4x^3-21x^2+4x=0\\ x\left(4x^2-21x+4\right)=0\end{array}$

Simplify the above expression further.

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

And

  $4x^2-21x+4=0$

To solve the above equation, use the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

  $\begin{array}{c}x=\frac{-\left(-21\right)\pm \sqrt{{\left(-21\right)}^2-4\left(4\right)\left(4\right)}}{2\left(4\right)}\\ =\frac{21\pm \sqrt{441-64}}{8}\\ =\frac{21\pm \sqrt{377}}{8}\end{array}$

Solve for x .

  $\begin{array}{c}x=\frac{21+\sqrt{377}}{8}\\ x=5.052\end{array}$

And

  $\begin{array}{l}x=\frac{21-\sqrt{377}}{8}\\ x=0.197\end{array}$

Thus, the roots of the expression are $x=0$ , 5.052 and 0.197.

Hence, the required horizontal tangents of the curve are $x=0$ , $x=5.052$ and $x=0.197$ .