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Please help with differentiation!

  • Thread starter toasted
  • Start date
1. The problem statement, all variables and given/known data

The function f(x) is defined as f(x)= -2(x+2)(x-1)^2 on the open interval (-3,3).

a. Let g(x) be defined as g(x)= abs(f(x)) in the open interval (-3,3). determine the coordinate(s) of the relative maxima of g(x) in the open interval. Explain your reasoning.

b. For what values of g'(x) not defined? Explain your reasoning.

c. Find all values of x for which g(x) is concave down. Explain your reasoning.



I was absent the past few days and so I missed how to do these types of problems in class. Could someone show me how to get through it, so I can complete my other homework.
 
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1. The problem statement, all variables and given/known data

The function f(x) is defined as f(x)= -2(x+2)(x-1)^2 on the open interval (-3,3).

a. Let g(x) be defined as g(x)= abs(f(x)) in the open interval (-3,3). determine the coordinate(s) of the relative maxima of g(x) in the open interval. Explain your reasoning.

b. For what values of g'(x) not defined? Explain your reasoning.

c. Find all values of x for which g(x) is concave down. Explain your reasoning.



I was absent the past few days and so I missed how to do these types of problems in class. Could someone show me how to get through it, so I can complete my other homework.
Can you sketch the graph of y = f(x)? From that it's pretty easy to get the graph of y = g(x); namely any part of the graph of f that is below the x-axis will be reflected across the x-axis. Let's start with that.
 
Can you sketch the graph of y = f(x)? From that it's pretty easy to get the graph of y = g(x); namely any part of the graph of f that is below the x-axis will be reflected across the x-axis. Let's start with that.
Ok so I understand part a then, because it makes sense that the f(x) values can no longer be negative. So for the next part I know that a function isn't defined when there's a sharp turn, or vertical tangent. But how do I apply that to the problem?
 
32,519
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At the places where the graph of f crosses the x-axis, when you reflect the negative portion back across the axis, you're going to have cusps (sharp corners). To represent this idea in text, at a place where the graph looks like this-- \ -- after you reflect the lower part, the graph will look sort of like this-- V-- and you have a cusp.

In particular, this is going to happen at x = -2. It's not going to happen at x = 1.
 

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