How to Use the Critical Value Method With Polynomials
The critical value method is an approach to graphing that uses calculus. By seeing when a function and its derivative equal zero, you can more easily graph the general shape of the function. The derivative is the slope of a curve at a point, also known as the instantaneous rate of change. When a function's slope is horizontal, its derivative there is zero. Therefore you can find peaks and valleys of a polynomial by finding where its first derivative is zero. This may seem intimidating, but calculus uses powerful methods that are much easier to use than the geometric methods taught in pre-calculus classes to find the same critical points. When the second derivative is zero, and is of opposite sign on either side of this critical point, then the concavity of the polynomial has changed. For example, if the graph of an upside-down cup and a right-side-up cup join, their meeting point is a critical point where the graph is neither concave down nor up.
Other People Are Reading
Instructions
-
-
1
Solve for the zeroes of the polynomial and mark them on your graph paper along the x-axis. This is where the polynomial crosses the x-axis.
For example, suppose your function of x is the polynomial f(x) = 2x^3 + 3x^2 - 8x + 3, where the caret ^ indicates exponentiation. This factors out to (2x-1) (x+3) (x-1). This equals zero at x = ½, -3 and 1.
-
2
Take the first derivative of f(x). Use the fact that the derivative of x^n is n*x^(n-1) for non-zero n, where the asterisk * indicates multiplication.
Continuing with the above example, the first derivative is f`(x) = 6x^2 + 6x - 8. Setting this to zero gives -0.5 +/- √57/6, where +/- indicates two answers, as in the quadratic formula.
-
-
3
Take the second derivative of f(x) and set it to zero. Solve for the zeroes.
Continuing with the above example, f``(x) = 12x+6 = 0. This gives x = -1/2. So there's an inflection point in the concavity at x = -1/2.
-
4
Draw lines through all the critical points.
Continuing with the above example, because large negative x values make f(x) negative, the first zero at x=-3 is intersected with a positive slope, coming from the lower left, moving up to the right. So draw a mark through that point on the x-axis with a short line indicating that positive slope. Similar reasoning leads to a negative-slope mark at x=1/2, and a positive-slope mark at x =1.
Draw a short horizontal line at x= -0.5 - √57/6 = -1.76, with a height f(-1.76)= -3.12. Now you can connect the two critical points at x=-3 and x=1/2 by passing the curve between them through the point (-1.76,-3.12). Perform a similar operation with the other critical point of the first derivative, at x=-0.5 + √57/6.
Draw the polynomial straight at x=-1/2, since that's were the polynomial is neither concave-up nor concave-down.
-
1
