Math 229
Curve Sketching Example
Let f(t) = t2
t2−1. Before we even begin, we should notice that f(t) is not defined for t=±1.
A. Find the intercepts.
y-intercept: f(0) = 02
02−1=0
−1= 0
x-intercepts: For a fraction to be zero, its numerator must be zero (and its denominator must be non-zero)
Here, if t2= 0, then t= 0, so there is only one intercept, the point (0,0) which is both an x-intercept and a y-intercept.
B. Finding Increasing/Decreasing Intervals and Relative Extrema Using f′(t).
f′(t) = 2t(t2−1) −t2(2t)
(t2−1)2=2t3−2t−2t3
(t2−1)2=−2t
(t2−1)2
Critical numbers:
Notice that f′(t) is undefined when t2−1 = 0 or when t=±1.
Also notice that f′(t) = 0 when t= 0.
Analyze the sign of f′(t):
+_
+_
x=0 x=1
x=−1
2−2 1/2−1/2
Therefore, f(x) is increasing on the intervals: (−∞,−1) ∪(−1,0)
Similarly, f(x) is decreasing on the intervals: (0,1) ∪(1,∞)
Classify Local Extrema:
Notice that f(0) is defined, and f′(t) goes from positive to negative at t= 0, so there is a local maximum when t= 0. The
value of this maximum is f(0) = 0, so the local maximum occurs at the point (0,0). This is the only local extremum.
C. Find Concavity and Inflection Points Using f′′(t).
f′′(t) = −2(t2−1)2−(−2t)(2)(t2−1)(2t)
(t2−1)4=(t2−1)[−2(t2−1) + (2t)(2)(2t)]
(t2−1)4=(t2−1)[−2t2+ 2 + 8t2]
(t2−1)4=(6t2+ 2)
(t2−1)3
To find the key values for the second derivative, notice that f′′(t) is undefined when t=±1 and that f′′(t) is textitnever
zero.
Sign testing diagram for f′′(t):
++
_
x=1
x=−1
2−2 0
Therefore f(x) is concave up on the intervals (−∞,−1) ∪(1,∞) and concave down on the interval (−1,1).
Notice that there are no inflection points, since the function is undefined at t=±1, and these are the only places where f(t)
changes concavity.
Combined Sign Chart:
