I believe it is the 3rd one.....I looked at it graphically, but I do not know exactly how to find the answer otherwise.....I'll think about it for a while and if I can come up with the REASON for the answer , I will re-post.... sorry ...
If you know how to do the derivative, it is where the slope goes from positive to negative.....
(or as ROM points out below....IF you know how to do the second derivative, it is where this = 0 )
I thhhink ur right ElectricPavlov. Cuz when I put the graph into www.desmos.com/calculator, the graph looked like it started its growth rate from the - side to the positive side. I'm just not sure if I'm right.
But I'll wait for your post :)
I think ROM will get it.....I used an online derivative calculator to find the result.....he can probably do it in his head !
Probably... he seems to be typing some massive MLA essay ;)
ElectricPavlov, what result did you get with your derivative calc?
Here is the funtion and the graph of the derivative and the line x = ln4/(.2)
\(\text{you need to find out where the 2nd derivative of }f \text{ is positive}\\ \text{It's the 3rd choice}\)
.hm this is hard
if a function f(x) is mapping x to y, then the inverse function f(x) maps y back to x
\(y=\frac{15}{1+4e^{-0.2x}}\)
interchange the variables x and y
\(x=\frac{15}{1+4e^{-0.2y}}\) solve for y
\(y=-5\ln \left(\frac{-x+15}{4x}\right)\)
\(-5\ln \left(\frac{-x+15}{4x}\right)\)
find the domain of each inverse function
domain of \(v\begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:
the domain of a function is the set of input or argument values for which the function is real and define
find positive values for log \(0
\(\log _af\left(x\right)\quad \Rightarrow \quad \:f\left(x\right)>0\)
\(\frac{-x+15}{4x}>0\)
multiply both sides by 4
\(\frac{4\left(-x+15\right)}{4x}>0\cdot \:4\)
simplify:
\(\frac{-x+15}{x}>0 \)
x<0 | X=0 | 0 | x=15 | x>15 | |
-x+15 | + | + | + | 0 | _ |
x | - | 0 | + | + | + |
-x+15/x | - | undefined | + | 0 | - |
OMG i cant do this anymore
srry , thats all i know
We aren't looking for the inverse function f^-1 we are speaking of the derivative of the function f '