Chapter 3.3, Problem 7E

To determine

To calculate: The horizontal tangents of the curve.

Answer to Problem 7E

The required horizontal tangents of the curve are $x=1$ and $x=\frac{1}{3}$ .

Explanation of Solution

Given information:

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

Formula used:

Power rule:

  $\frac{d}{dx}{x}^n=n{x}^{n-1}$

Calculation:

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

Use the power rule on the above curve.

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

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

  $3x^2-4x+1=0$

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

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

Simplify the above expression further.

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

And

  $\begin{array}{c}3x-1=0\\ 3x=1\\ x=\frac{1}{3}\end{array}$

Thus, the roots of the expression are $x=1,\frac{1}{3}$ .

Hence, the required horizontal tangents of the curve are $x=1$ and $x=\frac{1}{3}$ .