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EXAMPLE 4
Graphing the derivative given the graph of a polynomial function
Given the graph of a polynomial function y f 1x2, graph y f ¿1x2.
8
y
6
–3
–2
4
2
–1 0
–2
y = f(x)
x
1 2 3
–4
–6
–8
Solution
A polynomial function f is continuous for all values of x in the domain of f. The
derivative of f, f ¿, is also continuous for all values of x in the domain of f.
6
y
4 y = f(x)
2
x
–1 0
–2
1 2
–4
–6
4
y
2 y = f’(x)
x
–1 0
–2
1 2
–4
–6
–8
To graph y f ¿1x2 using the graph of y f 1x2, first determine the slopes of the
tangent lines, f ¿1x i 2, at certain x-values, x i . These x-values include zeros, critical
numbers, and numbers in each interval where f is increasing or decreasing. Then
plot the corresponding ordered pairs on a graph. Draw a smooth curve through
these points to complete the graph.
The given graph has a local minimum at (0, 1) and a local maximum at (1, 2). At
these points, the tangents are horizontal. Therefore, f ¿102 0 and f ¿112 0.
At x 1 , which is halfway between x 0 and x 1, the slope of the tangent is
2
2
about . So f ¿Q 1 2 R 2 3
3
The function f 1x2 is decreasing when f ¿1x2 6 0. The tangent lines show that
f ¿1x2 6 0 when x 6 0 and when x 7 1. Similarly, f 1x2 is increasing when
f ¿1x2 7 0. The tangent lines show that f ¿1x2 7 0 when 0 6 x 6 1.
The shape of the graph of f 1x2 suggests that f 1x2 is a cubic polynomial with a
negative leading coefficient. Assume that this is true. The derivative, f ¿1x2, may be
a quadratic function with a negative leading coefficient. If it is, the graph of f ¿1x2
is a parabola that opens down.
Plot (0, 0), (1, 0), and Q1 2
, on the graph of f ¿1x2. The graph of f ¿1x2 is a parabola
2 3 R
that opens down and passes through these points.
NEL
CHAPTER 4 177