Part 2: Now that we know what kinds of information we are looking for we need to

Question

Part 2: Now that we know what kinds of information we are looking for we need to discover how to find it. For each question below set the function equal to zero and solve for x. Then, plug those values back into f(x) and locate where the resulting point(s) exists on the graph at the top of page one. For questions 2 and 3 see if you can find a way to use the point(s) you found to break the graph into intervals that will help you fill out the tables. Also, using either or both of your textbook and lecture notes find the first and second derivative tests and determine how they can help you.

What information about its graph can we determine from the function itself? Hint: Think back to our discussion of limits. Use the function above f(x)=x^3-8x^2-20x to find this information.



What information about its graph can we determine from the 1st derivative of the function? The first derivative of our function is f^' (x)=3x^2-16x-20.


Interval
X-value in interval
Sign of f%u2019(x) at chosen value
Conclusion about increasing/decreasing behavior

What information about its graph can we determine from the 2nd derivative of the function? The second derivative of our function is f"(x)=6x-16.


Interval
x-value in interval
Sign of f%u201D(x) at chosen value
Conclusion about concavity.


What is the 1st derivative test and how can it help you answer part of question 2? In other words, how are we using the increasing/decreasing behavior of a function to locate the minima and maxima of that function?


What is the 2nd derivative test and how can it help you answer part of question 2? In other words, how are we using the concavity of the function to locate the minima and maxima of the function?

Explanation / Answer

f(x) = x^3-8x^2-20x = 0 => x(x^2 - 8x - 20) = 0 => x(x - 10)(x + 2) = 0 => x = -2, 0, 10 are the roots. The graph of f(x) cuts x-axis at -2, 0 and 10. f'(x) = 3x^2 - 16x - 20. At x = -2, f'(-2) = 3(-2)^2 - 16(-2) - 20 = 12 + 32 - 20 = 24 > 0 => slope at x = -2 is +ve => f(x) is increasing at x = -2. At x = 0, f'(0) = 3(0)^2 - 16(0) - 20 = - 20 < 0 => slope at x = 0 is -ve => f(x) is decreasing at x = 0. At x = 10, f'(10) = 3(10)^2 - 16(10) - 20 = 300 - 160 - 20 = 120 > 0 => slope at x = 10 is +ve => f(x) is increasing at x = 10. Let's look at f''(x) = 6x - 16. At x = -2, f''(-2) = 6(-2) - 16 = -12 - 16 = -28 < 0 => increase in the slope at x = -2 is -ve => slope of f(x) is decreasing at x = -2. At x = 0, f''(0) = 6(0) - 16 = - 16 < 0 => increase in the slope at x = 0 is -ve => slope of f(x) is decreasing at x = 0. At x = 10, f''(10) = 6(10) - 16 = 60 - 16 = 44 > 0 => increase in the slope at x = 10 is +ve => slope of f(x) is increasing at x = 10.

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