1. ## Trying again

Still unsure of how to find the inflection points with an e^x problem...

Take this problem:
f(t) = 1 - e^-0.03t
f'(t) = -e^-.03t + (-.03t)
f'(t) = -0.03te^-0.03t

It seems logical that if t=0 then f'(t) =0, but is there a way to solve it without guessing?

f"(t) = -.03t(e^-.03t)(-.03)
f"(t) = .0009te^-.03t

In other words, can you set .0009t = e^-.03t and solve? or do you just pick some points near 0 and plot?

2. Hello, confusedagain!

Still unsure of how to find the inflection points with an e^x problem...

Take this problem:
$\displaystyle f(t) \:= \:1 - e^{-0.03t}$
$\displaystyle f'(t) \:= \:-e^{-0.03t} + (-0.03t)$ ?

The derivative of: .$\displaystyle y \:=\:e^u$ is: .$\displaystyle y' \:=\:e^u\cdot u'$

In baby-talk: the exponential function itself times the derivative of the exponent.

If $\displaystyle f(t)\:=\:1 - e^{-0.03t}$

. . then: .$\displaystyle f'(t)\:=\:-e^{-0.03t}(-0.03) \:=\:0.03e^{-0.03t}$

. . then: .$\displaystyle f''(t)\:=\:0.03e^{-0.03t}(-0.03) \:=\:-0.0009e^{-0.03t}$

To find inflection points, solve $\displaystyle f''(t) = 0$

So we have: .$\displaystyle -0.0009e^{-0.03t} \:=\:0$

. . Divide by -0.0009: .$\displaystyle e^{-0.003t} \:=\:0$

. . Multiply both sides by $\displaystyle e^{0.03t}\!:\;\;1 \:=\:0$ . . . . What?

Therefore, there are no inflection points.

3. Originally Posted by confusedagain
Still unsure of how to find the inflection points with an e^x problem...

Take this problem:
f(t) = 1 - e^-0.03t
f'(t) = -e^-.03t + (-.03t)
f'(t) = -0.03te^-0.03t

It seems logical that if t=0 then f'(t) =0, but is there a way to solve it without guessing?

f"(t) = -.03t(e^-.03t)(-.03)
f"(t) = .0009te^-.03t

In other words, can you set .0009t = e^-.03t and solve? or do you just pick some points near 0 and plot?
Hello,

I assume that your function is:
$\displaystyle f(t)=1-e^{-0.03t}$
To calculate the first derivative use chain rule:

$\displaystyle f'(t) = -e^{-0.03t} \cdot (-0.03) = 0.03 \cdot e^{-0.03t}$

Both factors are positiv thus the product is positiv too, that means $\displaystyle f'(t) \neq 0$

To calculate the second drivative use chain rule again:
$\displaystyle f''(t) = 0.03 \cdot e^{-0.03t} \cdot (-0.03) = -0.0009 \cdot e^{-0.03t}$
Both factors have always different signs thus the product is negative for all t.

The function don't have any extreme points or inflection points if the domain is $\displaystyle \mathbb{R}$