Question

How are these calculus questions different?

How are a) and b) different? What should be done for each one?

1. Let f(x) = x^2 - 4x - 5

a) Find f '(3) using the definition f a = lim h->0 f(a+h) - f(a)/h

b) Find the slope of the tangent line to the graph of f(x) at x=3

(I thought slope of the tangent line at x=3 was the same as finding the derivative of 3?)

Answer 1

Both answers are 2.

Explanation:

In the case of question (a), you are being asked to find the derivative `f'(x)`, evaluated at `x = 3` using the limit definition of a derivative. It's typically the first step students encounter when learning about derivatives and what they mean.

In this case, the limit definition essentially describes the concept of using secant lines to approximate the slope of a tangent to a function, while gradually reducing the "spacing" between the two `x` points used to make the secant line slope approximation. See the following graph for an example of this:

In the graph, the solid green curve is `f(x)`. The dotted lines are secant lines made through three sets of additional points on `f(x)` "away from" where `f(3)` is located. The red line is made using the "farthest away" point (3.75, -5.9375), while the green line is made using the "closer" point (3.5, -6.75), while the black dotted line is made using the "closest" point (3.25, -7.4375). The solid blue line is the true tangent line. Notice how the slopes of each dotted line become more and more similar to the true tangent as the additional point gets closer on the curve to (3, -8) (the value of `f(3)`). The limit definition of the derivative expresses this in mathematical terms.

The slope is traditionally phrased as "rise over run", meaning the amount the function changes in the y-direction (`Deltay`) divided by the amount the function changes in the x-direction (`Deltax`). If we choose some `x` distance "ahead of" 3, say `h`, then we can evaluate the function `f(x)` at that location of `3+h` by calculating `f(3+h)`. If we also evaluate `f(3)`, we can figure out the change in y (`Deltay`) by finding the difference between those values, or `Deltay = f(3+h) - f(3)`.

Likewise, we know the difference between the two x coordinates is simply `h` because that's how far to the right we moved from our starting point. If you don't believe that, then calculate `Deltax = (3+h)-3 = h`.

This gives us an expression for the slope of the line connecting `(3, -8)` and `(3+h,f(3+h))`:

`slope = (Deltay)/(Deltax)=(f(3+h)-f(3))/h`

The "magic" comes from then using a limit to mathematically "see" what happens when we make the separation `h` get smaller and smaller and smaller, until eventually we make the difference `h` disappear. Visually, you can see that in the picture above. Mathematically, we calculate that using a limit and allowing `h` to go to 0, or `h->0`:

`f'(3) = lim_ {h->0} (f(3+h) - f(3))/h`

`= lim_ {h->0} (((3+h)^2-4(3+h)-5) - (3^2-4(3)-5))/h`

`= lim_ {h->0} ((9+6h+h^2-12-4h-5) - (9-12-5))/h`

`= lim_ {h->0} (-8+2h+h^2-(-8))/h`

`= lim_ {h->0} (2h+h^2)/h = lim_ {h->0} (cancel(h)(2+h))/cancel(h)`

`= lim_ {h->0} (2+h) = 2`

What this limit says is if you let the two points get closer and closer together, the slope will approach the value 2. (If you are patient, you can see that in my picture above, as the slopes of the dotted lines are 2.75, 2.5, and 2.25 - all values heading towards 2.)

In question (b), you are correct in noting that the slope of the tangent line to `f(x)` at `x = 3` would equal the derivative `f'(3)`, which was found in part (a). I'm not certain that anything different is expected; usually these exercises are meant to have you derive the understanding first, and then use that understanding to answer the basic question which comes next.

Answer 2

Check your textbook/teacher's definitions.

Explanation:

Sometimes (for example in James Stewart's text) the slope of the tangent line at `x=a` is defined by

`m = lim_(xrarra)(f(x)-f(a))/(x-a)` if the limit exists.

While the derivative of `f` at `a` is

`f'(a) = lim_(hrarr0)(f(a+h)-f(a))/h` if the limit exists.

Theorem: The two limits above either both exist and are equal or neither exists.

So the numerical answer will be the same either way.