Calculus Volume 1

Chapter 4 | Applications of Derivatives

393

of these points. For example, let’s choose x =−2, x =0, and x =4 as test points. Interval Test Point Sign of f ′ ( x ) =3 ( x −3 )( x +1 ) at Test Point

Conclusion

(−∞, −1)

(+)(−)(−) = +

x =−2

f is increasing.

(−1, 3)

(+)(−)(+) = −

x =0

f is decreasing.

(3, ∞)

(+)(+)(+) = +

x =4

f is increasing.

Step 3. Since f ′ switches sign from positive to negative as x increases through 1, f has a local maximum at x =−1. Since f ′ switches sign from negative to positive as x increases through 3, f has a local minimum at x =3. These analytical results agree with the following graph.

Figure 4.32 The function f has a maximum at x =−1 and a minimum at x =3

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