1. ## Miserable... tangent

Hey, guys. I'm new to the whole calculus branch of mathematics (we've just started) and I'm utterly confused about some concepts. Initially, we had started working finding the slope of a tangent to a particular point on a curve via m= f(a+h)-(a)/ h and working it by subbing in the given point etc... now in another chapter, the equation for calculating the derivative function is given, f ' (x) = lim h-->0 f(x+h)-f(x)/ h .. I'm curios to know what's the difference.. just the a is replace by a different variable.

2. I do not understand the question. But you need to take the limit, because subsitutiing that value at that point is just not define.

Post #1

3. Find an equation of the tangent to the graph of f(x)= 1/x where x= 2.

Find an equation of the tangent to the graph of f(x)= 1/x where x= 2.
$\displaystyle f(x)=\frac{1}{x}=x^{-1}$
$\displaystyle \frac{dy}{dx}=-1 \times x^{-1-1}=\frac{-1}{x^2}$

Now when you plug in x=2 you get $\displaystyle \frac{-1}{4}$

This is the gradient of both the curve and the tangent line at x = 2. However, the tangent line has a constant gradient, so you have

$\displaystyle y-y_0=\frac{-1}{4}(x-x_0)$ as the equation of the tangent line.

From $\displaystyle f(x)=\frac{1}{x}$, plugging in $\displaystyle x=2$ you get $\displaystyle y=\frac{1}{2}$ (finding this y coordinate [which is on the tangent line as well] gives us enough info to complete the equation)

=> $\displaystyle y-\frac{1}{2}=\frac{-1}{4}(x-2)$

=> $\displaystyle y=1-\frac{x}{4}$