Maximum and Minimum Problems in Exponential Functions

[dnartS]

I'm having a hard time with these two questions for an assignment in grade 12 advanced functions..

1) Find the maximum and minimum values for each exponential growth or decay equation on the given interval.
i) y=100(0.85)^t, for 0 (less than or equal to) t (less than or equal to) 5
ii) y = 35(1.15)^x, for 0 (less than or equal to) x (lessthan or equal to) 10

(This one has me puzzled, I am not sure where to even start)

and

2) Use an algebraic strategy to verify that the point given for each function is either a maximum or minimum.
i) f(x) = x^3 - 3x ; (-1, 2)

The relevant equation for number 2) would be difference quotient:
f(x) - f(x)+h
h

(h is 0.0001 which is a very close number to 0)so

f(h) = x^3 - 3x

= [(-1)^3 - 3(-1)] - [(-1)^3 - 3(-1) + 0.0001]
0.0001

= 2 - 2.0001
0.0001

= 0

:. The point is a maximum because the slope of the tangent is 0
Now I'm not sure if this is correct, but the answer is supposed to be maximum. Any help would be appreciated because this is really tricky.

 

[LCKurtz]

You said The relevant equation for number 2) would be difference quotient:
\frac {f(x) - f(x)+h} {h}

That isn't correct. It should be

\frac {f(x+h) - f(x)} {h}

 

[Mark44]

2) Use an algebraic strategy to verify that the point given for each function is either a maximum or minimum.
i) f(x) = x^3 - 3x ; (-1, 2)
The relevant equation for number 2) would be difference quotient:
f(x) - f(x)+h
h

(h is 0.0001 which is a very close number to 0)

so

f(h) = x^3 - 3x

= [(-1)^3 - 3(-1)] - [(-1)^3 - 3(-1) + 0.0001]
0.0001

= 2 - 2.0001
0.0001

= 0

:. The point is a maximum because the slope of the tangent is 0

There are several errors in the work above. As LCKurtz already pointed out, the difference quotient is (f(x + h) - f(x))/h.

You have f(h) = x³ - 3x, which is incorrect. f(h) = h³ - 3h.

In your next line, you have
= [(-1)^3 - 3(-1)] - [(-1)^3 - 3(-1) + 0.0001]
0.0001

The 0.0001 in the 2nd line is supposed to indicate division. What you are apparently trying to do is to evaluate (f(-1 + h) - f(-1))/h, but as already noted, you are using an incorrect expression for the difference quotient.

In the next line, you have
= 2 - 2.0001
0.0001

= 0
It looks like you are attempting to evaluate (2 - 2.0001)/0.0001 = -0.0001/0.0001 = -1, not 0.

As it turns out, f'(-1) = 0, but your work doesn't support this conclusion. Having found that the slope of the tangent line at x = -1 is zero doesn't necessarily mean that the point is a maximum. It could just as easily have been a minimum, and for some functions it can be neither. To show that there is a maximum at x = -1 you could show that the slope of the tangent line is positive for values of x near -1 but to the left of it, and that the slope of the tangent line is negative for values of x near -1 but to the right of it.

Finally, and this is only a comment, not an error, 0.0001 is really not all that close to 0. I can come up with numbers that are 100 or 1000 times closer to zero.

 

[dnartS]

How could I prove that the tangents before the maximum are positive and that the tangents after the maximum are negative?

(The difference quotient was taught to me that way by my teacher, but after reading the textbook you're right, but it doesn't make a difference because it would still be -1)


I'm not sure how I could support the conclusion of verifying this point is a max.

edit: A smaller number such as 0.000000000000000000001 would be of course more ideal but just to make it simpler I wrote 0.0001

 

[Mark44]

How could I prove that the tangents before the maximum are positive and that the tangents after the maximum are negative?

How about looking at these difference quotients?
(f(-1 - h) - f(-1))/h, where h > 0, and for several values of h. These would give you the average rates of change for values to the left of x = -1.

And for values to the right of x = -1, you could evaluate this difference quotient:
(f(-1 + h) - f(-1))/h, where h > 0, and for several values of h.

(The difference quotient was taught to me that way by my teacher, but after reading the textbook you're right, but it doesn't make a difference because it would still be -1)

If this is the difference quotient taught to you by your teacher -
\frac {f(x) - f(x)+h} {h}
- I don't have much confidence in your teacher. This one always gives a value of 1, no matter what the function is, as long as f is defined at x, and h is not zero. And it does matter, because the real difference quotient in your problem produces a value of -1, which is different from 1.

The fact that

I'm not sure how I could support the conclusion of verifying this point is a max.

edit: A smaller number such as 0.000000000000000000001 would be of course more ideal but just to make it simpler I wrote 0.0001

I didn't have any problem with your choice of a number for h. My quibble was with you saying that 0.0001 was "very close" to zero.

 

[HallsofIvy]

How could I prove that the tangents before the maximum are positive and that the tangents after the maximum are negative?

You mean that the slopes of the tangent lines are positive and negative.

If a gives a maximum that means f(a) is larger than other, near by, values. The function must be increasing for x below a and decreasing for x above a. Thus positive slope before a and negative slope after.

(The difference quotient was taught to me that way by my teacher, but after reading the textbook you're right, but it doesn't make a difference because it would still be -1)

You don't think it might be possible that you misunderstood or misremembered what your teacher said?

I'm not sure how I could support the conclusion of verifying this point is a max.

edit: A smaller number such as 0.000000000000000000001 would be of course more ideal but just to make it simpler I wrote 0.0001

But the point is that no matter HOW small a number you take there will always be numbers much smaller than that. Even if I get a quotient of 0 taking a number within 0.00000000000000000000000000000000000001 of 2, that will not prove the function does not suddenly drop very far down even closer.