Chapter 14.4, Problem 19AYU

To determine

To find: The slope of the tangent to the function $y = f\left(x\right)$ at the indicated point.

Answer to Problem 19AYU

13

Explanation of Solution

Given:

$f\left(x\right) = x^3 + x$

Point: $\left(2, 10\right)$

Formula used:

The slope of the tangent to the curve $y = f\left(x\right)$ is the derivative of $f\left(x\right)$ w.r.to $x$ ie $\frac{dy}{dx}$.

Calculation:

Here, $y = f\left(x\right) = x^3 + x$

Slope of the tangent is given by;

$\frac{dy}{dx} = \frac{d}{dx}\left(x^3 + x\right)$

$\frac{dy}{dx} = \frac{d}{dx}\left(x^3\right) + \frac{d}{dx}\left(x\right)$

$\frac{dy}{dx} = 3x^2 + 1$

At the point $\left(2, 10\right)$; $\frac{dy}{dx} = 3\left(2^2\right) + 1 = 12 + 1 = 13$

Hence the slope of the tangent to the curve $y = f\left(x\right) = x^3 + x$ at $\left(2, 10\right)$ is 13.

The graph is as follows: