Finding the Tangent Line at a Point

Tangent line = a straight line that barely passes through a curve.

1. Tangent lines are equal to zero when the line change from increase to decrease or the opposite way around.

This image shows the the line changing from increase to decrease at (-1,2) and (1,2) and their slope is equal to zero because it is a straight line with a rise of zero.

Slope and Equation of a Tangent Line

A line tangent to a curve is one that only touches the curve at only one point. In order to find the equation of a line tangent to a curve at a certain point, you have to find the slope of the curve at that point, which requires calculus. You can then write the equation of the tangent line in point-slope form. The tangent line has to be moving in same direction and speed of the point in that curve.

(Video) Finding the Linearization at a Point

by PatrickJMT

Here's a video by patrickJMT showing you how to find a tangent line approximation, or linearization, of a function.

When you're asked to find a linearization of something what they're really just saying is to find a tangent line at that point. So if we're told to find the linearization of: $$f(x)=\frac{1}{\sqrt{7+x}}\ \text{ at }\ x=0$$ What it's asking for is the equation of the line tangent to $f(x)$ at $x-0$.

So we know that $x=0$, therefore the $y$-coordinate will be $f(0)$. $$y=\frac{1}{\sqrt{7+\color{red}{(0)}}}=\frac{1}{\sqrt7}$$

Now that we have an $x$- and a $y$-coordinate we can begin constructing the equation for our tangent line. $$\begin{align} y-y_1&=m(x-x_1)\\ y-\frac{1}{\sqrt7}&=m(x-0) \end{align}$$

To find the slope, $m$, we need to take the derivative of our function and plug in our point, $x=0$. $$\begin{align} f(x)&=\frac{1}{\sqrt{7+x}}\\ &=(7+x)^{-\frac12}\\ f'(x)&=-\frac12(7+x)^{-\frac32}\\ &=-\frac{1}{2\sqrt{(7+x)^3}}\\ f'(0)&=-\frac{1}{2\sqrt{(7+\color{red}{0})^3}}\\ &=-\frac{1}{2\sqrt{7^3}} \end{align}$$

We can plug this back into our tangent equation for the slope. $$\begin{align} y-\frac{1}{\sqrt7}&=-\frac{1}{2\sqrt{7^3}}(x-0)\\ y-\frac{1}{\sqrt7}&=-\frac{1}{2\sqrt{7^2\cdot7}}(x)\\ y-\frac{1}{\sqrt7}&=-\frac{x}{2\cdot7\sqrt{7}}\\ y-\frac{1}{\sqrt7}&=-\frac{x}{14\sqrt{7}}\\ y&=\frac{1}{\sqrt7}-\frac{x}{14\sqrt{7}}\\ y&=\frac{1}{\sqrt7}\bigg(1-\frac{1}{14}x\bigg) \end{align}$$

(Videos) Finding a Linear Approximation

by PatrickJMT

Here's a video by patrickJMT where he goes through an example of finding the linearization of a function at a point.

In these problems we're just asked to find the linear approximation at the point. All that is is the tangent line at that point. It's not asking us to approximate some unknown value like in this example.

The problem asks for you to find a linear approximation to the function $f(x)=2^x$ at the point $x=3$.

We use the same method that we have been using to find the tangent line.

1. Using point-slope form, construct a line, adding in the given point.
2. Take the derivative of the function, then plug in the point, to find the slope for the tangent function.

To begin we construct our tangent line equation. $$\begin{align} y-y_0&=m(x-x_0)\\ y-\color{red}{(2^3)}&=m(x-\color{red}{(3)})\\ y-8&=m(x-3) \end{align}$$

Now we take the derivative and evaluate it at our point. $$\begin{align} f(x)&=2^x\\ f'(x)&=2^x\cdot\ln2\\ f'(3)&=2^3\cdot\ln2\\ m&=8\ln2 \end{align}$$

Plugging this back into our tangent line equation, we get our linearization. $$\begin{align} y-8&=m(x-3)\\ y-8&=8\ln2(x-3)\\ y-8&=(8\ln2)x-24\ln2\\ y&=(8\ln2)x-24\ln2+8 \end{align}$$

Here's another video with the second example.