Assignment #2:

Explore the relationship between the graph of a function and the graph of the derivative of the function. Given a function f(x), what is the equation of the tangent line at any given point. Using Graphing Calculator 2.2, graph f(x) and the tangent line. (Hint: let n represent "any given point.")

The best way that I can see to start this exploration is to start with a function that we all know well, f(x)=x^2.

Here is the graph of f(x)=x^2. No surprise, it is a parabola. Now, we know that the derivative of a function gives us the slope of the tangent line to f(x) at any given point x.

So, in our case, the derivative f'(x)=2x. Let's look at this graph.

What this shows is that the slope of the tangent line of the original function, at any given point x, will be the corresponding y value.

So at point x=1, the slope of the tangent line will be 2. Easily confirmed by plugging x=1 into f'(x), 2(1)=2=m.

So now, since we know the slope of the tangent line, we can graph the tangent line by using the point slope equation of a line. At x=1, the point (1,1^2) is (1,1),

and the slope m, is 2(1)=2. So the point slope form is y-y1=m(x-x1), or in this case y-1=2(x-1) which is equal to y=2x-1.

Here, the blue graph is that of our tangent line. To see this
graph at any point n, __click here__.

So, in general, if f(x)=x^2, then f'(x)=2x, so at any arbitrary point n, f(n)=n^2 and f'(n)=2n.

Here, 2n is the slope of the tangent line to our original function at any arbitrary point.

What does all of this tell us? Obviously we need to figure out what the formula will be for any function.

Well let's see, in this case the equation of our tangent line is y-f(x)=f'(x)(x-x1).

So, in general, would it be true that y-f(n)=f'(n)(x-n) gives us the graph of the tangent line at any point n?

Let's try another example to see if this does work.

f(x)=x^2+x

f'(x)=2x+1

So, I claim that the equation of the tangent line will be in the form y-f(n)=f'(n)(x-n), which in our case is y-(n^2+n)=(2n+1)(x-n).

Simplifying this equation, we get y=(2n+1)x-n^2.

The graph of these three equations look like so:

Here the blue graph is that of our tangent line. To see this
graph in action, __click here__.

So, this shows that our equation, y-f(n)=f'(n)(x-n) will work for any function.

In conclusion, we can graph a function, and we can graph the derivative of that function. Moreover, we know that the derivative gives us the slope of the tangent line to the original function. Using this knowledge, we can deduce that the equation of the tangent line to our function at any given point, using the point slope form will be

y-f(n)=f'(n)(x-n).

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