Concept explainers
To sketch: a possible graph of f gives that it is continuous with domain [-3, 2] and
f (-3) =0.
Answer to Problem 35RE
Explanation of Solution
Given information: The graph of f’ is given.
Calculation:
From the graph of f’, know that f has a max at x = -2 since f’ switches from positive to negative and f has a min at x = 1 since f switches from negative to positive. Since f’ is linear for -3< x <0, f should be quadratic for these values. Since f ‘ is quadratic for 0 < x < 2, f should be cubic for these values. It is given that f (-3 )= 0 so plot a point at (-3, 0). Since f’ > 0 for -3 < x < -2, draw an increasing quadratic curve between -3 and -2 so that x = -2 will be the vertex. Since there is a max at x =-2 and f’< 0 for -2 < x <0, draw a quadratic curve that is decreasing from -2 to 0. Form the graph of f’, to the left of f ‘(0), it is about -2 so draw this part of the curve to have a tangent slope of about -2 at x =0 . From x= 0 to x = 1, f’ < 0 so draw a decreasing cubic curve starting from the point at x =0 from the quadratic portion since the curve starting from the point at x = 0 from the quadratic portion since the curve must be continuous. From the graph of f’ , to the right of f’(0)it is about -1 so draw this portion to have a tangent slope of about -1 around x = 0. Since there is a min at x= 1 and f’ > 0 for x=1 to x =2, draw an increasing cubic curve for these values.
Chapter 5 Solutions
Calculus: Graphical, Numerical, Algebraic: Solutions Manual
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