Question

If f(x) = eˣ and g(x) = $log_{e} x$, then show that fog = gof and find f⁻¹ and g⁻¹.

Answer 1

Answer:

fog = gof

Step-by-step explanation:

In the attachments I have answered this problem.

Concept:

If f and g are two functions such that

fog = gof = I where I is identity function then inverse of f = g and inverse of g = f.

See the attachment for detailed solution.

Answer 2

Step by step Explanation:

As we know that fog(x) = f(g(x))

$fog(x) = {e}^{ log_{e}(x) } = x \\ \\ gof(x) = log_{e}( {e}^{x} ) = x$
since log and exponential cancels each other.

Finding f inverse:

$y = f(x) \\ \\ y = {e}^{x} \\ \\ log \: y = log \: {e}^{x} \\ \\ x = log \: y \\ \\ x = f(y) \\ \\ {f}^{ - 1}( x) = log_{e}\: x$

Finding g inverse:


$y = g(x) \\ \\ y = log_{e}(x) \\ \\ {e}^{y} = {e}^{ log_{e}(x) } \\ \\ x = {e}^{y} \\ \\ {g}^{ - 1} (x) = {e}^{x}$
Hope it helps you