MHB How Can I Simplify This Tangent Line Equation?

[Teh]

View attachment 6121View attachment 6119
stuck on how to simplify this?

 

[MarkFL]

We are given the function:

$$f(x)=\frac{3}{x}$$

Using the given definition, we are to compute:

$$f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\lim_{x\to a}\frac{\dfrac{3}{x}-\dfrac{3}{a}}{x-a}=\lim_{x\to a}\frac{3a-3x}{ax(x-a)}=-\lim_{x\to a}\frac{3(x-a)}{ax(x-a)}=-\lim_{x\to a}\frac{3}{ax}=-\frac{3}{a^2}$$

And so the tangent line at the point $(b,f(b))$ would be given by (using the point-slope formula):

$$y=-\frac{3}{b^2}(x-b)+\frac{3}{b}=-\frac{3}{b^2}x+\frac{6}{b}$$

Here's an interactive graph to show you the functions and some of its tangent lines...you can use the "slider" for $b$ to see the tangent line for different values of $b$.

[desmos=-5,-5,5,5]y=\frac{3}{x};y=-\frac{3}{b^2}x+\frac{6}{b};b=1[/desmos]

In this problem, we are given $b=4$...so what is the tangent line?

 

[Teh]

We are given the function:

$$f(x)=\frac{3}{x}$$

Using the given definition, we are to compute:

$$f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\lim_{x\to a}\frac{\dfrac{3}{x}-\dfrac{3}{a}}{x-a}=\lim_{x\to a}\frac{3a-3x}{ax(x-a)}=-\lim_{x\to a}\frac{3(x-a)}{ax(x-a)}=-\lim_{x\to a}\frac{3}{ax}=-\frac{3}{a^2}$$

And so the tangent line at the point $(b,f(b))$ would be given by (using the point-slope formula):

$$y=-\frac{3}{b^2}(x-b)+\frac{3}{b}=-\frac{3}{b^2}x+\frac{6}{b}$$

Here's an interactive graph to show you the functions and some of its tangent lines...you can use the "slider" for $b$ to see the tangent line for different values of $b$.
In this problem, we are given $b=4$...so what is the tangent line?

the tangent line will be 3/2

 

[MarkFL]

the tangent line will be 3/2

The will be the $y$-intercept, but not the line itself. :D

 

[karush]

$$\displaystyle f(x)= -\frac{3}{16} x+\frac{3}{2}$$