1. ## Functions

Functions f and g are defined for x E R by

f:x --->e^x
g:x--->2x-3

(i) Solve the equation fg(x)=7

Function h is defined as gf
(ii)Express h in terms of x and state its range
(iii)Express h inverse in terms of x.

2. ## Re: Functions

(i)x=(ln7+3)/2...its easy. take ln both sides.
(ii)h(x)=2e^x-3
its minimum value will be when the term 2e^x is minimum since 3 is constant ..2e^x can attain a minimum zero(just attain) at x=-infinity.And its maximum value will be infinity.therefore range is (-3,infinity)
(iii)for finding h inverse take log again..
Hope it helps....

3. ## Re: Functions

Note that:

$\displaystyle f \circ g(x) = f(g(x))$

So, in this case, that means:

$\displaystyle f \circ g(x) = e^{g(x)} = e^{2x-3} = (e^{2x})(e^{-3}) = \frac{(e^x)^2}{e^3} = 7$

$\displaystyle e^x = \sqrt{7e^3}$, and taking logs, we get:

$\displaystyle x = \log(\sqrt{7e^3}) = \log((\sqrt{7})(e^{\frac{3}{2}}))$

$\displaystyle = \log(\sqrt{7}) + \log(e^{\frac{3}{2}}) = \log(\sqrt{7}) + \frac{3}{2} = \frac{\log(7) + 3}{2}$

Similarly,

$\displaystyle g \circ f(x) = 2(e^x) - 3$.

Since $\displaystyle e^x$ ranges from (0,∞), it's clear that $\displaystyle 2(e^x)$ also ranges from (0,∞) and thus the range of $\displaystyle g \circ f$ is (-3,∞).

Consequently, $\displaystyle h^{-1} = f^{-1} \circ g^{-1}$ is only defined on (-3,∞). Can you figure out what it is, in terms of x?

4. ## Re: Functions

i solved it maybe it is : (ln(x+3)/2)=h-1(x)
here domain of h-1(x) is (-3,infinity) same as the range ..