Finding inverse derivative for f(x)=e^3x + 2e^x - 5 at a=-2

[moham_87]

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

 

[PrudensOptimus]

$$f(x) = e^{3x} + 2e^x - 5$$

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

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

$$y = f^{-1}(x)$$
$$y' = 1/(10x)$$

 

[himanshu121]

hey PrudensOptimus What are u doing. These properties of ln are not in the league