instantaneous rate of change question help please

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Summer5526

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hello! if anyone could please help me with this question I've been stuck on it for 1 hour anyone who does is an absolute angel.

the question: when is the instantaneous rate of change zero for the function: A(x)= (0.5x^2+x+2) / x
I have not yet learned calculus so I have to do this without finding the derivative!

thank you so much!
 
Summer5526 said:
hello! if anyone could please help me with this question I've been stuck on it for 1 hour anyone who does is an absolute angel.

the question: when is the instantaneous rate of change zero for the function: A(x)= (0.5x^2+x+2) / x
I have not yet learned calculus so I have to do this without finding the derivative!

thank you so much!
Click to expand...
What class is this? Did the teacher say that you could not use derivatives?
 
lev888 said:
What class is this? Did the teacher say that you could not use derivatives?
Click to expand...
it’s advanced functions so we have not learned calculus yet, i’m assuming the teacher wants us to use the instantaneous rate of change formula which is f(x+h)- f(x) / h and equate it to zero but i’m not too sure how to do so with the type formula given
 
Here is a hint. The instantaneous rate of change is at a point where the the tangent line is zero. As Harry_the_cat suggested, draw the graph.
 
The change in f, as x goes from a to a+ h, is f(a+ h)- f(a). The rate of change is \(\displaystyle \frac{f(a+h)- f(a)}{h}\).

The "instantaneous rate of change" is the limit as h goes to 0.

Here \(\displaystyle f(a)= \frac{0.5a^2+ a+ 2}{a}\) and \(\displaystyle f(a+ h)= \frac{0.5(a+ h)^2+ a+ h+ 2}{a+ h}= \frac{0.5a^2+ ah+ 0.5h^2+ a+ h+ 2}{a+h}\) so \(\displaystyle f(a+ h)- f(a)= \frac{0.5a^2+ ah+ 0.5h^2+ a+ h+ 2}{a+h}- \frac{0.5a^2+ a+ 2}{a}\).

To subtract we need to get a common denominator.
\(\displaystyle f(a+h)- f(a)= \frac{(0.5a^2+ ah+ 0.5h^2+ a+ h+ 2)a}{a(a+h)}- \frac{(0.5a^2+ a+ 2)(a+ h}{a(a+ h}\)
\(\displaystyle = \frac{0.5a^3+ a^2h+ 0.5ah^2+ a^2+ 2a- 0.5a^3-0.5a^2h- a^2- ah- 2a- 2h}{a(a+ h)}\)
\(\displaystyle = \frac{(0.5a^2h+ 0.5ah^2- ah- 2h}{a(a+ h)}= \frac{(0.5a^2+ 0.5ah- a- 2)h}{a(a+h)}\)

So the "difference quotient} is
\(\displaystyle \frac{f(x+h)- f(x)}{h}= \frac{0.5a^2+ 0.5ah- a- 2)h}{a(a+h)h}= \frac{0.5a^2+ 0.5ah- a- 2}{a(a+h)}\)

Now take the limit of that as h goes to 0.
(Check my algebra. I'm not at all sure I haven't mis-written.)
 
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