Hi again, here is my problem and my efforts

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The discussion centers on finding the inverse function of f(x) = e3x + 2ex - 5 for x ≥ 0, specifically determining the derivative of the inverse function at a = -2. The user successfully proved that f(x) is increasing by calculating f'(x), but encountered difficulty solving the equation e3x + 2ex - 3 = 0 to find the corresponding x-value. The user attempted to manipulate the equation using logarithmic properties but faced challenges in applying them correctly.

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i wish i could find a solution for that poblem

it is about inverse function:
Show that the function has an inverse function, and find [(d/dx)f^-1(x))] x=a for the given number a

f(x)=e^3x + 2e^x - 5 , x>=0 , a=-2

Here i proved that the function has an inverse by finding f'(x), and found that it is increasing at the given interval.

then f(x)= -2 to find "x", and then substitute in the theory:
g(a)=1/[f'(g(a))]

the problem here is that i can't find "x"
where, e^3x + 2e^x - 3 = 0

please i need help, any efforts will be appreciated
 
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[tex]f(x) = e^{3x} + 2e^x - 5[/tex]

to find, f^-1(x), change x to y. find y.

[tex]x = e^{3y} + 2e^y - 5[/tex]
[tex]lnx = 3y + ln 2e^y - ln5[/tex]
[tex]lnx = 4y + ln 0.4[/tex]
[tex]ln 2.5x = 4y[/tex]
[tex]y = ln2.5x/4[/tex]

[tex]y = f^{-1}(x)[/tex]
[tex]y' = 1/(10x)[/tex]
 
hey PrudensOptimus What are u doing. These properties of ln are not in the league
 

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