Is Every Transformation of Utility Functions Monotonic?

[djalap]

Homework Statement

Suppose that the utility functions u(x,y) are related to v(x,y) = f(u(x,y)). In each case below, select "Yes" if the function f is an increasing, monotonic transformation and "No" if it is not.

Homework Equations

A differentiable equation f(u) is an increasing function of u if its derivative is positive.

The Attempt at a Solution

f(u) = 3.141592u (YES)
f(u) = 5000-23u (YES)
f(u) = u-100000 (YES)
f(u) = log(base 10)u (NO)
f(u) = -e^-u (NO)
f(u) = 1/u (NO)
f(u) = -1/u (YES)

My answers are in parenthesis. Are they right? If not, can you explain briefly why?

 

[Mark44]

Homework Statement

Suppose that the utility functions u(x,y) are related to v(x,y) = f(u(x,y)). In each case below, select "Yes" if the function f is an increasing, monotonic transformation and "No" if it is not.

Homework Equations

A differentiable equation f(u) is an increasing function of u if its derivative is positive.

The Attempt at a Solution

f(u) = 3.141592u (YES)
f(u) = 5000-23u (YES)

Above, check the derivative. Is it positive?

f(u) = u-100000 (YES)
f(u) = log(base 10)u (NO)

Above, Why do you think this?

f(u) = -e^-u (NO)

Above, check the derivative. Is it negative?

f(u) = 1/u (NO)
f(u) = -1/u (YES)

The two above are a little tricky. For the first, the derivative is not defined for u = 0, so, for example, even though 1/u is decreasing on each half of its domain it is not decreasing overall. -1 < 1, but 1/-1 < 1/1, which should not be the case for a decreasing function.

My answers are in parenthesis. Are they right? If not, can you explain briefly why?

 

[djalap]

Above, check the derivative. Is it positive?

Derivative of 3.141592u = 3.141592 = positive
Derivative of 5000-23u = -23 = negative. so its no

Above, Why do you think this?

Derivative of u-100000 = 1 = positive
Deriviative of log(base 10)u = 1/ln(10) = 2.3 = Positive

Above, check the derivative. Is it negative?

derivative of e = 0? so then, is that positive or negative?

The two above are a little tricky. For the first, the derivative is not defined for u = 0, so, for example, even though 1/u is decreasing on each half of its domain it is not decreasing overall. -1 < 1, but 1/-1 < 1/1, which should not be the case for a decreasing function.

these 2 are correct?

New answers =
Yes
No
Yes
Yes
IDK? (NO?)
No
Yes

 

[Mark44]

Deriviative of log(base 10)u = 1/ln(10)

d/du(log u) is not a constant.

derivative of e = 0?

Yes, but that's not the function - it is -e^-u. What is d/du(e^u)? What is d/du(e^-u)? Take another look at the differentation rules for log x, ln x, e^x.

 

[djalap]

so d/du (log u) = 1/(u LN 10) = no?

d/du (-e^-u) = e^-u = no

 

[Mark44]

so d/du (log u) = 1/(u LN 10) = no?

You have the derivative right, but isn't 1/(u ln 10) > 0?

d/du (-e^-u) = e^-u = no

Again, the derivative is right, but isn't e^-u > 0?

 

[djalap]

You have the derivative right, but isn't 1/(u ln 10) > 0?

Again, the derivative is right, but isn't e^-u > 0?

With U in the function, can't U be negative, making the answer negative?

 

[Mark44]

The domain for log u is {u | u > 0}. This has an effect on the values of 1/(u ln 10).
The domain for e^-u is all real numbers, and the range is the same as for e^u. (The graphs are different, of course.)

 

[djalap]

The domain for log u is {u | u > 0}. This has an effect on the values of 1/(u ln 10).
The domain for e^-u is all real numbers, and the range is the same as for e^u. (The graphs are different, of course.)

So they are both positive.

Because e^-u is always positive.

Right?

 

[Mark44]

"They" being the derivatives - yes.