G(x) is twice differentiable where g(4)=12 and g(5)=18. g(6)=?

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Homework Help Overview

The discussion revolves around a twice differentiable function g with specific properties: g'(x) > 0 and g''(x) > 0 for all real numbers x, along with given values g(4) = 12 and g(5) = 18. Participants are exploring potential values for g(6) based on these conditions.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to understand the implications of the positivity of the first and second derivatives. There is discussion about the linear approximation leading to a value of 24 for g(6) and questioning why this contradicts the conditions given.

Discussion Status

Some participants have offered guidance on interpreting the conditions of the problem, suggesting that the increasing and concave properties of g imply that g(6) must be greater than 24. The possibility of using an exponential function is also mentioned as a potential approach to satisfy the conditions.

Contextual Notes

There is a recognition that the problem is framed within a multiple-choice context, which may influence the reasoning and approaches taken by participants. The discussion reflects on the constraints of the problem without resolving them.

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Homework Statement



Let g be a twice differentiable function with g'(x)>0 and g''(x)>0 for all real numbers x, such that g(4)=12 and g(5)=18. Of the following, which is a possible value for g(6)?

a. 15
b. 18
c. 21
d. 24
e. 27

Answer: e. 27


Homework Equations





The Attempt at a Solution



I guess the first question is the first line of the question. "Let g be a twice-differentiable function with g'(x)>0 and g''(x)>0" means the first and second derivative of g(x) is positive, right?

They gave me g(4)=12 and g(5)=18. Therefore, I can find the function by plugging it into y-y1 = m(x-x1, find m, and then find b.

12-18 = m(4-5)
-6 = m(-1)
m=6

12 = 6(4) + b
b = -12

y= 6x-12​

Since they want g(6), I plugged in 6 for x.

y= 6(6)-12
y= 24​

Though that is answer d, it is incorrect.

I guess my problem might come from the first sentence. They told me g'(x) and g''(x) is positive, but I don't know how that helps me.
 
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Almost there! Your solution y=24 is the straight line solution with slope=6. In other words, g'(x)=6. Since this is a straight line, the second derivative is zero. However, the problem tells you that g'(x) and g''(x) are positive. This means g(6) must be greater than 24. The only possible choice is (e) 27.
 
g' and g'' give you some idea about the shape of the graph. If g'(x) > 0, the graph is increasing. If g''(x) > 0, that gives you information about the concavity of the graph, whether the graph is concave up or concave down.
 
g''(x)>0 means there must be a place where it is concave up, right?

y= 6x-12 is a straight line.
y'=6 is also a straight line
y'' does not exist? which means there is no concavity which means g''(x) is not greater than 0.

Therefore when I plug in 6, I get 24. But because of the conditions above, 24 CANNOT be the answer. And since y'' must be positive, the answers cannot be anything below 24, leaving me with the only answer choice left, (e) 27?
 
If y = 6x - 12, then y' = 6, and y'' = 0, so yes, it does exist, and there is no concavity.
 
Your function must be monotone increasing and concave for all x. So an exponential function comes into mind. As you have two constraints, try g(x) = a exp( bx). Solving this for your two x values will lead to a very simple result with indeed g(6) = 27.

Erich
 
esc1729 said:
Your function must be monotone increasing and concave for all x. So an exponential function comes into mind.
Or not. This problem can be solved without having to solve for a specific function.
esc1729 said:
As you have two constraints, try g(x) = a exp( bx). Solving this for your two x values will lead to a very simple result with indeed g(6) = 27.

Erich
 
Yeah, I've just realized. It's quite some time since I'm out of school and we had no multiple choice tests then ...
 
No, I didn't either, but a problem like this is good for emphasizing concepts at a high level without getting bogged down in computations.
 

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