At what points does the graph of the function y(x) = x^3 - 4 x^2 + 3x + 5 have a horizontal tangent line?
The derivative of this function is
- y ' (x) = ( x^3) ' - 4 ( 2 x^2 ) ' + 6 (x) ' - (2) ' = ( 3 x^3) + 6 ( 2x) + 3 - 0, so that
- y ' (x) = 3 x^2 - 8 x + 3.
The derivative y ' (x) is the instantaneous rate of change of y with respect to x. This rate is identified with the slope of the line tangent to the graph of the function.
A horizontal tangent line has slope 0, since between any two distinct points the rise is 0 and the run is nonzero. Any time 0 is divided by a nonzero number, the result is 0.
So the function has a horizontal tangent line whenever we have the condition
- y ' (x) = 0, or
- 3 x^2 - 8 x + 3 = 0.
Using the quadratic formula we easily find that this equation has two solutions:
- x = (- (-8) - sqrt( (-8)^2 - 4 * 3 * 3) / (2 * 3) and x = (- (-8) +sqrt( (-8)^2 - 4 * 3 * 3) / (2 * 3) . These solutions simplify to
- x = ( 8 - 2 sqrt(7) ) / 6 and x = ( 8 + 2 sqrt(7) ) / 6 or
- x = 4/3 - sqrt(7) / 3 and x = 4/3 + sqrt(7) / 3. The approximate values of these solutions are
- x = 2.22 and x = .45
These are the x coordinates of the points at which the tangent line is horizontal. The corresponding y coordinates are y(2.22) = 2.89 and y(.45) = 5.63.
On what intervals of the x axis will the graph of y(x) vs. x be increasing, and on what intervals will it be decreasing?
y(x) is increasing if the rate of change of y with respect to x is positive; this corresponds to a graph whose tangent line at every point has a positive slope, i.e., to a graph which goes 'uphill' as we move from left to right.
The tangent line has positive slope whenever y ' (x) > 0.
So to find where the graph is increasing we need to solve the inequality
- y'(x) > 0, or
- 3 x^2 - 8 x + 3 > 0.
The function 3 x^2 - 8 x + 3 is continuous; its graph has no 'breaks' or 'jumps'. As x changes the value of 3 x^2 - 8 x + 3 changes, and small changes in this quantity can be assured by making sufficiently small change in the value of x.
The only way the quantity 3 x^2 - 8 x + 3 can change from negative at one value of x to positive at another value of x, or from positive to negative, is to take the value 0 at some point in between. That is, you can't get from negative to positive without going through 0.
By the full statement of the quadratic formula, the only points at which 3 x^2 - 8 x + 3 takes the value 0 are at x = 4/3 - sqrt(7) / 3 and x = 4/3 + sqrt(7) / 3, approximately x = 2.22 and x = .45.
We are led to the following conclusions:
- There is no point between approximately x = .45 and x = 2.22 at which the value of 3 x^2 - 8 x + 3 is zero. Therefore this quantity is either positive on the entire interval between these points, or negative on the entire interval.
- We can therefore determine which is which by evaluating the function at any point in the interval. We pick a point at which the function is easy to evaluate: x = 1. We find that if x = 1, 3 x^2 - 8 x + 3 = -2. We conclude that 3 x^2 - 8 x + 3 is negative on the entire interval between .45 and 2.22.
- The same reasoning tells us that 3 x^2 - 8 x + 3 cannot be 0 at any point to the right of x = 2.22 (the only points at which the quantity can be 0 are at x = .45 and x = 2.22, and .45 is not to the right of 2.22). So if we evaluate the quantity at any point to the right of x = 2.22 the result tell us the sign of the quantity on the entire interval.
- x = 10 is a nice round number in the interval, and it's easy to evaluate 3 x^2 - 8 x + 3 for x = 10. We get 300 - 80 + 3 = 223. This number is positive, so 3 x^2 - 8 x + 3 is positive everywhere to the right of x = 2.22.
- We use the same reasoning for the interval to the left of x = .45. Since x = 0 is to the left of .45, and since is it very easy to evaluate 3 x^2 - 8 x + 3 for x = 0, we do so. We obtain the value 3, which is positive.
- Thus 3 x^2 - 8 x + 3 is positive for all values of x to the left of x = .45.
- In summary we find that y'(x) = 3 x^2 - 8 x + 3 is positive on the x-interval (-infinity, .45), negative on the x-interval (.45, 2.22) and positive on the x-interval (2.22, infinity).
Since y(x) is increasing where y ' (x) is positive and decreasing where y ' (x) is negative we conclude that the graph of y(x) is increasing on the intervals (-infinity, .45) and (2.22, infinity), and decreasing on the interval (.45, 2.22).
The graph has horizontal tangent lines at the x = .45 and x = 2.22 points.
We arrive at the following description of the graph:
The graph increases as we move from the left, until it peaks at the point (.45, 5.63).
The graph then decreases, reaching a 'valley' at the point (2.22, 2.89).
The graph then increases as we move to the right.