Chapter 14.4, Problem 12AYU

To determine

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

Answer to Problem 12AYU

$-2$

Explanation of Solution

Given:

$f\left(x\right) = 3 - x^2$

Point: $\left(1, 2\right)$

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, $y = f\left(x\right) = 3 - x^2$

Slope of the tangent is given by;

$\frac{dy}{dx} = \frac{d}{dx}\left(3 - x^2\right)$

$\frac{dy}{dx} = \frac{d}{dx}\left(3\right) - \frac{d}{dx}\left(x^2\right)$

$\frac{dy}{dx} = 0 - 2x$

$\frac{dy}{dx} = -2x$

At the point $\left(1, 2\right)$; $\frac{dy}{dx} = -2\left(1\right) = -2$

Hence the slope of the tangent to the curve $y = f\left(x\right) = 3 - x^2$ at $\left(1, 2\right)$ is $-2$.

The graph is as follows: