Two applications

We can now put this idea to good use. Consider the following graph of a function $f:[-10, 10]\rightarrow(-\infty,\infty)$.

The problem, called an optimization problem is the find the value of x which gives the largest value of f(x). The method of solution is to construct a sieve, that is, a rule which eliminates most values $x\in[-10, 10]$from consideration.

The idea for this sieve is that at if x = a gives the largest value of f(a), then the tangent line at a (if it exists) will have slope equal to , since the rate of change of f at x=a will have to be .

To understand why, consider that if h < 0 then the slope of the chord from (a+h,f(a+h)) to (a,f(a)) cannot be negative, and if h > 0, the slope of the chord cannot be positive. Hence, as h approaches , the only possible value for the rate of change is a number which is neither positive nor negative, namely .

Here is simple example demonstrating this. We pick this simple example because it is possible to verify that our approach yields the correct answer without using calclus at all. We take $f:(-\infty,\infty)\rightarrow(-\infty,\infty)$,f(x) = -x² + 6x + 10. We know that the graph of y = f(x) is a parabola that opens down and that intersects the x-axis at $x= 3\pm\sqrt{19}$, so the maximum value is midway between the x- intercepts, namely, at x=3, and the maximum value is 19. Alternatively, we can complete the square, and observe that -x² + 6x +10 = -(x-3)² + 19 which will be 19 at x=3.

What does our new method suggest? Well, since the rule for f will have a rate of change for every choice of x, we are looking for places where the rate of change is . We compute f'(x) = -1(2)x^2-1 + 6 = -2x+6, so the only place where the rate of change is is at x=3, in accordance with our earlier analysis.

Here is another application, finding the zeroes of an expression. Consider the function $f:(-\infty,\infty)\rightarrow(-\infty,\infty)$, f(x) = x⁵ + x -1. For which x is f(x) equal to . Unlike the situation for polynomials of degrees 1, 2, 3 and 4, there is no general rule for finding the roots of polynomials of degrees 5 or higher. All we know is that the number of roots is not more than the degree of the polynomial. It is pretty easy to see that if x < 0 then f(x) < 0 and if $x \geq 1$ then $f(x) \geq 0$. This suggests we look at a new function, $g:[0,1]\rightarrow(-\infty,\infty)$ with rule g(x) = x⁵ + x - 1. The benefit in considering g is that we only have to look at values of x in a much smaller set. Now the problem is that we have no idea how many zeroes g has. The graph of g could have any of the following shapes, since we do know that g(0) = - 1 and g(1) = 1.

How to decide? Notice that if there is more than one zero, the tangent line of g should be horizontal somewhere. Well, g'(x) = 5x⁴ + 1 which is never if $x\in [0, 1]$. Note that knowing the domain is crucial. If x could be a complex number, then there are four different values of x making g'(x) equal to .

So now we know there is one zero. How do we estimate it? One way is bisection: Compute f(1/2) = -15/32. Since this value is negative, the zero lies between 1/2 and 1. Now compute f(3/4) < 0, so the zero lies between 3/4 and 1. Next, we compute f(7/8) > 0 and so the zero lies between 3/4 and 7/8. On and on we go, and after 10 repetitions we have only determined the first three digits of the zero.

Here is another alternative, called Newton's method, which is much quicker. Since the easier rule to find a zero for is of the form t(x) = mx + b, and since the tangent line lies close to the graph, we try looking for x-intercepts of tangent lines to y = x⁵ + x + 1. If we start at x=1, we have that the slope of the tangent line is g'(1) = 5(1)⁴ + 1 = 6, and the tangent line has equation y - 1 = 6(x-1). The x-intercept is when y = 0,

so we need to solve 0-1 = 6(x-1), yielding x = 5/6.

Now we repeat the process, constructing the tangent line at x = 5/6: The slope is g'(5/6), the point of tangency is (5/6,g(5/6)) and an equation of the tangent line is y - g(5/6) = g'(5/6)(x-5/6), or

$$y - \frac{1829}{7776} = \frac{4421}{1296}x-\frac{22105}{7776},$$

which has x-intercept $x = 10138/13263 \approx 0.764$, and we can see from our graph that the true zero is smaller.

It will take the bisection method several more iterations just to reach the value we reached with Newton's method in two iterations.

For one last application of finding equations of tangent lines, complete the second problem on homework assignment three, concerning the optical properties of parabolas.