Average rate of change of an exponential function

[zhongkui]

When I graphed A(t) = 35e^-0.17t on my calculator, I ended up with the answers
t=2 is 24.912
t=4 is 17732
t=6 is 12.621
If the rate of change formula is A(b) - A(a)/ b-a and I wanted to find the average rate of change from t=4 to t=6 would t=4 be a or b.

 

[wjm11]

When I graphed A(t) = 35e^-0.17t on my calculator, I ended up with the answers
t=2 is 24.912
t=4 is 17732
t=6 is 12.621
If the rate of change formula is A(b) - A(a)/ b-a and I wanted to find the average rate of change from t=4 to t=6 would t=4 be a or b.

[A(b) - A(a)]/ [b-a] = [A(6) - A(4)]/ [6-4]

Note that the answer will turn out to be negative. For this exponential equation, we expect a negative slope/average rate of change, because the negative sign in the exponent indicates we have an exponential decay curve. The slope/average rate of change between any two points will be negative.

Also note, however, that had you switched values for "a" and "b", you'd still get a negative answer.

 

[Ishuda]

When I graphed A(t) = 35e^-0.17t on my calculator, I ended up with the answers
t=2 is 24.912
t=4 is 17732
t=6 is 12.621
If the rate of change formula is A(b) - A(a)/ b-a and I wanted to find the average rate of change from t=4 to t=6 would t=4 be a or b.

For $\displaystyle \frac{A(b) - A(a)}{b-a}$ standard procedure is, IMO, to have b > a. So, a would be 4 and b would be 6. In actuality, the answer is the same no matter which way you do it. If you interchange a and b, then both numerator and denominator would change signs and you would have the same value.

 

[HallsofIvy]

It doesn't matter if "b> a" or not:
$\displaystyle \frac{A(b)- A(a)}{b- a}= \frac{-(A(a)- A(b))}{-(a- b)}= \frac{A(a)- A(b)}{a- b}$.

Either choice for a and b, long as you are consistent, will give the same answer.