Average rate of change? How would I figure the following problem out? Find the average rate of change of g(x)=x^2+3x+7 from x=5 to x=9 My thought is that I would plug in 5 and 9 for the x values to get the y values. And the use the slope formula y^2−y^1/x^2−x^1.

aphathalo 2022-09-07 Answered
Average rate of change?
How would I figure the following problem out?
Find the average rate of change of g ( x ) = x 2 + 3 x + 7 from x=5 to x=9My thought is that I would plug in 5 and 9 for the x values to get the y values. And the use the slope formula y 2 y 1 x 2 x 1
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Answers (2)

Elliott Rollins
Answered 2022-09-08 Author has 8 answers
The Fundamental Theorem of Calculus says that a b f ( t )   d t = F ( b ) F ( a ) where the rate of change of F(t) with respect to t is f(t). The only thing that matters is where you start and where you end. Slight observation: if you have to make sure if you apply this to something like average speed as long as you used distance traveled rather than distance from start to end because speed is always a non-negative absolute value of velocity.In the case of this question, this says the average rate of change of g(x) (as opposed to the average change of g), is A v e r a g e   r a t e   o f   c h a n g e = g ( b ) g ( a ) b a = g ( 9 ) g ( 5 ) 9 5 = 9 2 + 3 9 + 7 ) ( 5 2 + 3 5 + 7 ) 9 5 = 115 47 47 = 17and this is exactly what a precalculus student should do.
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Gardiolo0j
Answered 2022-09-09 Author has 1 answers
Yes, your solution is correct.
Differentiating first is a detour.
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I assumed that the rate of change is the same as the gradient of the function, namely f. Calculating this gave me:
f = ( x y 2 z ) x i ^ + ( x y 2 z ) y j ^ + ( x y 2 z ) z k ^
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Hence, the function increases most rapidly in the x direction.
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asked 2022-08-16
Rates of change, compounding rates and exponentiationI have a very (apologies if stupidly) simple question about rates of change that has been bugging me for some time. I can't work out whether it relates to my misunderstanding what a rate of change is, to my misapplying the method for calculating a rate of change or something else. I'm hoping somebody on here can help.
For how I define a rate of change, take as an example a population of 1000 items (e.g. bacteria). I observe this population and after an hour I count the size of the population and see that it has increased by 10% (to 1100). I might hypothesise that the population is growing at the rate of 10% per hour, and if, an hour later, I see that it has grown by 10% again (to 1,210) then I might decide to conclude that it is growing at 10% per hour.
So, a rate of change of "proportion x per hour" means "after one hour the population will have changed by proportion x". If, after 1 hour, my population of bacteria was not 1,100, and if not 1,210 after 2 hours, that would mean that the rate of change was not 10% per hour.
First question: Is this a fair definition of a rate of change?
So far so good and it's easy to calculate the population after any given time using a compound interest-type formula.
But whenever I read about continuous change something odd seems to happen. Given that "grows at the rate of 10% per hour" means (i.e. is just another way of saying) "after 1 hour the original population will have increased by 10%", why do textbooks state that continuous change should be measured by the formula:
P = P 0 e r t
And then give the rate of change in a form where this seems to give the wrong answer (i.e. without adjusting it to account for the continuously compounded growth)? I've seen many texts and courses where 10% per day continuous growth is calculated as (for my above example, after 1 day):
1000 e 1 0.1 = 1105.17
This contradicts the definition of a rate of change expressed as "x per unit of time" stated above. If I was observing a population of 1000 bacteria and observed it grow to a population of 1105 after 1 hour I should surely conclude that it was growing at the rate of 10.5% per hour.
I can get the idea of a continuous rate just fine, and it's easy to produce a continuous rate of change that equates to a rate of 10% per day as defined above (that's just ln 1.1). But I struggle to see how a rate of change that means a population grows by 10.5% in an hours means it is growing at 10% per hour. That's like saying if I lend you money at 1% interest per month I'd be charging you 12% per year.
So what's wrong here? Have I got the wrong end of the stick with my definition of a rate of change, would most people interpret a population increase of 10.5% in an hour as a 10% per hour growth rate, or is something else amiss?
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What do instantaneous rates of change really represent?The derivative of f ( x ) is the value of the limit of the average rate of change of y with respect to x as the change in x approaches 0. This is the value, in other words, that the average rate of change approaches but NEVER hits.
This means that it is NOT the infintesimal rate of change of y with respect to d y / d x merely approaches the derivative's value. If the rate of change did actually achieve 0 change in x, you'd get 0/0 which is an indeterminant form.
So if the derivative is the literal rate of change at an exact instant -- a rate of change with an interval of 0, what does that actually tell you? Can a specific moment in time really have a rate of change? Is that rate of change ever even maintained, even at a specific instant?
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