Estimate the instantaneous rate of change in the equation 18999(0.93)^t when a = 5
Would it start like this?
(a+5)-(5)/h
18999(0.93)^(5+h) - 18999(0.93)^5 / h <<<not sure about this
13217^h - 13217 / h <<<<<not sure about this
Estimate the instantaneous rate of change in the equation 18999(0.93)^t when a = 5
Would it start like this?
(a+5)-(5)/h
18999(0.93)^(5+h) - 18999(0.93)^5 / h <<<not sure about this
13217^h - 13217 / h <<<<<not sure about this
Sure, the original function is that. But what's a? You say a = 5, but what does that mean? Do you mean to say that t = 5, i.e. you're investigating the instantaneous rate of change near where the function takes the value 5? If so, then the above is correct and there isn't a simplification of the last expression I wrote. So yeah, just plug in intsy-tintsy $\displaystyle h$.
You mean "when t= 5".
No, the "difference quotient" is $\displaystyle \frac{f(x+h)- f(x)}{h}$. You forgot the "f"! Also, you have the "a" where you should have "h"Would it start like this?
(a+5)-(5)/h
No, the first line above is correct but the second line is not. You cannot write $\displaystyle A(B^{5+h)}= (AB^5)^h$- that's just not true. What you want is A(B^{5+ h})= A(B^5)*B^h.18999(0.93)^(5+h) - 18999(0.93)^5 / h <<<not sure about this
13217^h - 13217 / h <<<<<not sure about this
Here, A= 18999 and B= .93 so [tex]AB^5= (18999)(0.6956883693)= 13217.3833283307 so that
$\displaystyle \frac{13217.3833283307(.93)^h- 13217.3833283307}h= 13217.3833283307\frac{(.93)^h- 1}{h}$.
But you are not finished. The "instantaneous rate of change" is the limit as h goes to 0. Assuming that limit exists, you can get a good approximation by using a value of h very close to 0, as ragnar suggests. It can, in fact, be shown that the limit does exist and is exactly equal to ln(.93).