English teacher studying for review

[Pool MgNoodle]

Hi friends, I'm an English teacher, and I never teach math, except I'll have to for an upcoming review (this is just a formality, if I had a real student with a real math problem, she'd be referred by me to a colleague capable of giving help.)

However I have a question about a problem that I might have to teach that I don't understand. Here's the problem:

If f(x) = (1/3)^x and a < b which of the following must be true?
a. f(a) > f(b)

To prove above correct answer, I have the following solution from a friend, but I don't understand the solution. Does anybody have a suggestion on the specific topic I might Google or video I might watch that could help with this?

f(a) = 1/3^a
f(b) = 1/3^b

If a < b
Step 1. f(a) + f(b) doesn't equal 3 (--> use any example to contradict)
Step 2. f(a) + 1/3 doesn't equal f(b) (--> use any example to contradict)
Step 3. f(a) = f(b) (--> contradict a < b)
Step 4. f(a) < f(b) (--> use any example to contradict)
Step 5. f(a) > f(b) (--> use any example to check)

I guess f(a) + f(b) cannot equal 3 because in either case, there'd be a fractional exponent applied to 1/3 and the numerator would stick to 1? Is that right?
After that step, why are the following proofs chosen to arrive at the answer? What is the thinking there? Couldn't we just as well come up with any giant number of numerical truths to write? But why those? And what is this language "use any example to contradict" what does the word contradiction have to do with solving the problem? I'm guessing this is some language used often to describe computer languages maybe? Finally, if anyone could give an idea on how better to understand the above problem and others like it, for example, the specific remedial lesson I need to take, I'd really appreciate that as well. Thank you!

 

[lev888]

Hi friends, I'm an English teacher, and I never teach math, except I'll have to for an upcoming review (this is just a formality, if I had a real student with a real math problem, she'd be referred by me to a colleague capable of giving help.)

However I have a question about a problem that I might have to teach that I don't understand. Here's the problem:

If f(x) = (1/3)^x and a < b which of the following must be true?
a. f(a) > f(b)

To prove above correct answer, I have the following solution from a friend, but I don't understand the solution. Does anybody have a suggestion on the specific topic I might Google or video I might watch that could help with this?

f(a) = 1/3^a
f(b) = 1/3^b

If a < b
Step 1. f(a) + f(b) doesn't equal 3 (--> use any example to contradict)
Step 2. f(a) + 1/3 doesn't equal f(b) (--> use any example to contradict)
Step 3. f(a) = f(b) (--> contradict a < b)
Step 4. f(a) < f(b) (--> use any example to contradict)
Step 5. f(a) > f(b) (--> use any example to check)

I guess f(a) + f(b) cannot equal 3 because in either case, there'd be a fractional exponent applied to 1/3 and the numerator would stick to 1? Is that right?
After that step, why are the following proofs chosen to arrive at the answer? What is the thinking there? Couldn't we just as well come up with any giant number of numerical truths to write? But why those? And what is this language "use any example to contradict" what does the word contradiction have to do with solving the problem? I'm guessing this is some language used often to describe computer languages maybe? Finally, if anyone could give an idea on how better to understand the above problem and others like it, for example, the specific remedial lesson I need to take, I'd really appreciate that as well. Thank you!

You need to post full text of the problem, it's pretty confusing to follow the proof with half of the problem missing.
You can show that the wrong answers are wrong by providing one example "to contradict" them.
To prove that a. is correct I would point out that (1/3)^x is a monotonically decreasing function - as x increases, y decreases.

 

[Otis]

… a problem that I might have to teach that I don't understand …

f(x) = (1/3)^x and a < b
f(a) > f(b) …

Hi PM. I'm not sure what level you're expected to lecture at. I'm thinking the "students" have already been taught basics of exponential functions, so you could begin by reminding them that exponential functions are 'well-behaved' functions (smooth and continuous). You could show two representative graphs, for students to visualize exponential change. Those curves have no turning points or gaps. As x increases, an exponential function has only one of two behaviors: f(x) keeps growing OR it's always getting smaller (exponential growth versus exponential decay). We can tell which, by looking at the base. In your exercise, function f is (1/3)^x. That function decreases continuously because the base (1/3) is less than 1. When a base is greater than 1, an exponential function increases continuously.

You could demonstrate, using positive integer values of x like {1,2,3} and reminding students that such exponents indicate the number of factors (1/3).

f(1) = 1/3
f(2) = (1/3)(1/3) = 1/9
f(3) = (1/3)(1/3)(1/3) = 1/27

The more factors we multiply (that is, as x increases), the larger the denominator gets while the numerator is fixed. In other words, the products decrease in value, as x increases.

Students ought to have learned also the definition of a negative exponent (it indicates the reciprocal of the product).

f(-3) = (3/1)(3/1)(3/1) = 27
f(-2) = (3/1)(3/1) = 9
f(-1) = (3/1) = 3

Tracing your finger (left to right) along a decreasing exponential graph, students can see that any f(a) will come before any f(b), so f(a) must always be larger than f(b).

?