1. 1 Example 4 Sketch the graph of the function k(x) = (x 2 -4) 4/5. Solution Observe that k is an even function, and its graph is symmetric with respect to the y-axis. I. Intercepts The x-intercepts occur when 0 = x 2 -4, i.e. when x=-2 and when x=2. The y-intercept occurs at II. Asymptotes The graph of k has no asymptotes. III. First Derivative By the chain rule, Since k / (x)>0 for –2 2, the function k is increasing there. Since k / (x)<0 for x<-2 and 0<x<2, the function k is decreasing there. We depict this information on a real number line.

2. 2 Note that k has three critical points: x=0 where the derivative is zero and x=-2, x=2 where the derivative does not exist. By the First Derivative Test x=-2 is a local minimum, x=0 is a local maximum and x=2 is a local minimum. IV. Vertical Tangents and Cusps Observe that at x=-2 the left derivative of k is -  and the right derivative is + , and k has a vertical cusp there. Similarly, at x=2 the left derivative of k is -  and the right derivative is + , and k also has a vertical cusp there. V. Concavity and Inflection Points By the quotient rule:

3. 3 VI. Sketch the graph We summarize our conclusions and sketch the graph of k. Since the concavity of k changes from up to down at there is an inflection point there. Since the concavity of k changes from down to up at there is an inflection point there. The denominator of k // (x) is always positive, so k // (x) has the same sign as its numerator. Hence k // (x) is positive for and is concave up there. k // (x) is negative for and is concave down there. We sketch this information on a number line.

4. 4 k is an even function x-intercepts: x=-2 and x=2 y-intercept: increasing: –2 2 decreasing: x<-2 or 0<x<2 local min: x=-2 and x=2 local max: x=0 vertical cusps: x=-2 and x=2 concave up: concave down: inflection points: k(x) =(x 2 - 4) 4/5