Chapter 3.3, Problem 38E

To determine

To find: The tangent to the curve for why $y=x^3+x$ at the points.

Answer to Problem 38E

The equation of the tangent is $y=4x+2$ .

Explanation of Solution

Given:

The given function is,

  $y=x^3+x$

The points where the slope is 4.

Calculation:

Consider the given question is,

  $y=x^3+x$

The slope of the tangent to the curve $y=f\left(x\right)$ is given by the formula is,

  $\frac{dy}{dx}=f^{\prime }\left(x\right)$

Then, the formula for the derivative is,

  $\frac{d{\left(x\right)}^{\prime \text{}\prime }}{dx}=n{\left(x\right)}^{n-1}$

Then,

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

The slope of the tangent is 4 then,

  $\begin{array}{l}3x^2+1=4\\ 3x^2=3\\ x^2=1\\ x=\pm 1\end{array}$

Consider the equation of the straight line is,

  $y=mx+c$

Then, from the equation of the tangent is,

  $\begin{array}{l}x=1\\ y=2\end{array}$

The equation of tangent at the point $\left(1,2\right)$ is,

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

Then, the point $\left(1,2\right)$ is $y=4x-2$ and the equation of the tangent at the point $\left(-1,-2\right)$ is

  $y=4x+2$ .