Question

Help with derivative of exponential function problem please!

Answer 1

Is it possible to solve for the minimum without the value for k?

Answer 2

Do you mean, is it possible to find a minimum for T(x) that is independent of k? I doubt it.

What did your derivative look like? \(\dfrac{dT}{dx}\)

Answer 3

The derivative is supposed to be...
but idk how to get it.

Answer 4

Well, generally, and you should drag through this until you get it...

\(\dfrac{dT}{dx} = -N\cdot p^{-x}\cdot \left(k\cdot \ln(p)+\dfrac{c\cdot \ln(p)}{x} + \dfrac{c}{x^{2}}\right)\)

We have N = 10

\(\dfrac{dT}{dx} = -10\cdot p^{-x}\cdot \left(k\cdot \ln(p)+\dfrac{c\cdot \ln(p)}{x} + \dfrac{c}{x^{2}}\right)\)

We have p = 0.9 and ln(p) = -0.10536

\(\dfrac{dT}{dx} = 10\cdot (0.9)^{-x}\cdot \left(k\cdot (0.10536)+\dfrac{c\cdot (0.10536)}{x} - \dfrac{c}{x^{2}}\right)\)

We have c = 1

\(\dfrac{dT}{dx} = 10\cdot (0.9)^{-x}\cdot \left(k\cdot (0.10536)+\dfrac{0.10536}{x} - \dfrac{1}{x^{2}}\right)\)

If we ALSO have k = 1, which we are not given, then we are done.

\(\dfrac{dT}{dx} = 10\cdot (0.9)^{-x}\cdot \left(0.10536+\dfrac{0.10536}{x} - \dfrac{1}{x^{2}}\right)\)

Not sure what to tell you about the missing assumption.

Answer 5

oh wow, well that was a lot simpler than i thought...
so just straight plugging in the numbers and assuming k = 1, yes?

Answer 6

You missed this part: "you should drag through this until you get it..."

Work out the derivative! The rest is substitution and a little algebra.

Answer 7

ooh okies thanks!