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3.1. USING DERIVATIVES TO IDENTIFY EXTREME VALUES
has neither a minimum nor a maximum at x = −1.
Activity 3.1.
Suppose that g(x) is a function continuous for every value of x 2 whose first
derivative is g
′ (x) =
(x + 4)(x − 1) 2
x − 2
. Further, assume that it is known that g has a
vertical asymptote at x = 2.
(a) Determine all critical numbers of g.
(b) By developing a carefully labeled first derivative sign chart, decide whether g
has as a local maximum, local minimum, or neither at each critical number.
(c) Does g have a global maximum? global minimum? Justify your claims.
(d) What is the value of lim
x→∞
g
′ (x)? What does the value of this limit tell you about
the long-term behavior of g?
(e) Sketch a possible graph of y = g(x).
⊳
The second derivative test
Recall that the second derivative of a function tells us several important things about the
behavior of the function itself. For instance, if f ′′ is positive on an interval, then we know
that f ′ is increasing on that interval and, consequently, that f is concave up, which also
tells us that throughout the interval the tangent line to y = f (x) lies below the curve at
every point. In this situation where we know that f ′ (p) = 0, it turns out that the sign of
the second derivative determines whether f has a local minimum or local maximum at
the critical number p.
In Figure 3.5, we see the four possibilities for a function f that has a critical number p
at which f ′ (p) = 0, provided f ′′ (p) is not zero on an interval including p (except possibly
at p). On either side of the critical number, f ′′ can be either positive or negative, and
hence f can be either concave up or concave down. In the first two graphs, f does not
change concavity at p, and in those situations, f has either a local minimum or local
maximum. In particular, if f ′ (p) = 0 and f ′′ (p) < 0, then we know f is concave down at
p with a horizontal tangent line, and this guarantees f has a local maximum there. This
fact, along with the corresponding statement for when f ′′ (p) is positive, is stated in the
3.1. USING DERIVATIVES TO IDENTIFY EXTREME VALUES
has neither a minimum nor a maximum at x = −1.
Activity 3.1.
Suppose that g(x) is a function continuous for every value of x 2 whose first
derivative is g
′ (x) =
(x + 4)(x − 1) 2
x − 2
. Further, assume that it is known that g has a
vertical asymptote at x = 2.
(a) Determine all critical numbers of g.
(b) By developing a carefully labeled first derivative sign chart, decide whether g
has as a local maximum, local minimum, or neither at each critical number.
(c) Does g have a global maximum? global minimum? Justify your claims.
(d) What is the value of lim
x→∞
g
′ (x)? What does the value of this limit tell you about
the long-term behavior of g?
(e) Sketch a possible graph of y = g(x).
⊳
The second derivative test
Recall that the second derivative of a function tells us several important things about the
behavior of the function itself. For instance, if f ′′ is positive on an interval, then we know
that f ′ is increasing on that interval and, consequently, that f is concave up, which also
tells us that throughout the interval the tangent line to y = f (x) lies below the curve at
every point. In this situation where we know that f ′ (p) = 0, it turns out that the sign of
the second derivative determines whether f has a local minimum or local maximum at
the critical number p.
In Figure 3.5, we see the four possibilities for a function f that has a critical number p
at which f ′ (p) = 0, provided f ′′ (p) is not zero on an interval including p (except possibly
at p). On either side of the critical number, f ′′ can be either positive or negative, and
hence f can be either concave up or concave down. In the first two graphs, f does not
change concavity at p, and in those situations, f has either a local minimum or local
maximum. In particular, if f ′ (p) = 0 and f ′′ (p) < 0, then we know f is concave down at
p with a horizontal tangent line, and this guarantees f has a local maximum there. This
fact, along with the corresponding statement for when f ′′ (p) is positive, is stated in the