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Chapter 4: Problem 37

Make a complete graph of the following functions. A graphing utility is useful in locating intercepts, local extreme values, and inflection points. $$f(x)=\frac{1}{3} x^{3}-2 x^{2}-5 x+2$$

Short Answer

Expert verified
Question: Sketch the graph of the function $$f(x) = \frac{1}{3}x^3 - 2x^2 - 5x + 2$$, and identify the x-intercepts, y-intercept, local maximum, local minimum, and inflection point. Answer: The x-intercepts are approximately (-2.44, 0), (0.775, 0), and (3.67, 0); the y-intercept is at (0, 2); there is a local maximum at x = -1; a local minimum at x = 5; and an inflection point at x = 2. (Use a graphing utility to sketch the graph including these points and features.)

Step by step solution

01

1. Find the x- and y-intercepts

To find the x-intercepts, we set the function equal to 0 and solve for x: $$0 = \frac{1}{3}x^3 - 2x^2 - 5x + 2$$ This is a cubic equation, and solving cubic equations analytically can be tough. So, we will use a graphing utility to find the approximate x-intercepts. To find the y-intercept, we set x=0: $$f(0) = \frac{1}{3}(0)^3 - 2(0)^2 - 5(0) + 2 = 2$$ So, the y-intercept is at the point (0, 2).
02

2. Find the local extreme values

To find the local extreme values, we need to find the critical points of the function. We do this by finding the first derivative and setting it equal to 0. $$f'(x) = x^2 - 4x - 5$$ Now, set f'(x) = 0 and solve for x: $$0 = x^2 - 4x - 5$$ Factor the equation: $$(x-5)(x+1) = 0$$ This gives us two critical points: x = 5 and x = -1. To classify these points as local minima or maxima, we use the second derivative test. First, find the second derivative: $$f''(x) = 2x - 4$$ Now, evaluate the second derivative at the critical points: $$f''(-1) = 2(-1) - 4 = -6$$ $$f''(5) = 2(5) - 4 = 6$$ Since f''(-1) < 0, we have a local maximum at x = -1. Since f''(5) > 0, we have a local minimum at x = 5.
03

3. Find the Inflection Points

To find the inflection points, first identify where the second derivative is equal to 0: $$0 = 2x - 4$$ Solve for x: $$x = 2$$ Now, since the second derivative changes sign around x=2, we have found an inflection point at $$x = 2$$.
04

4. Make the complete graph using the located values

By using a graphing utility, we are now able to sketch the graph of the function. Mark the following points on the graph : x-intercepts, y-intercepts, local maximum at x = -1, local minimum at x = 5, and inflection point at x = 2. Then, draw a smooth curve that connects these points and includes all identified features.

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