Exponential and Logarithmic Functions

The original function has an undefined slope at some point in its domain.

The original function only includes positive integers in its range.

The original function isn't one-to-one over its domain.

The original function contains no exponents.

Plug in values from one into another and check if it results in identity.

Replace 'x' with 'y' in both functions and solve simultaneously.

Plug in values from one into another and check if it results in zero.

Add both functions together to see if they cancel out.

It means that the output of the function will always be greater than the input.

It means the function can be reversed to find the original input from its output.

It means the function can only take positive values.

It means that the function is continuous and differentiable.

It remains non-invertible regardless of being injective or not after this change

It becomes discontinuous automatically at that point

There’s always an asymptote formed at that point in such cases

It may become invertible at that point if defined properly on its domain

The graph of the function has no breaks or holes

The function fails the Horizontal Line Test

The function is continuous on its domain

The function passes the Vertical Line Test

Zero

Negative infinity

Positive infinity

The limit does not exist.

$f^{-1}(y) = \frac{y - 2}{3}$

$f^{-1}(y) = \frac{2}{3}y - 1$

$f^{-1}(y) = \frac{3}{y + 2}$

$f^{-1}(y) = \frac{3}{y - 2}$

Continuous and one-to-one

Discontinuous and one-to-one

Continuous but not one-to-one

Discontinuous and not one-to-one

The distance $b$ is unrelated to the slope when comparing the function and its inverse.

The intercept $b$ of the function $j^{-1}(y)$ yields $j$ is the sum of the coefficients $a$ and the slope of the function $j$.

The coefficient of the function $j^{-1}(y)$ yields $j(yx)$ is the reciprocal of the coefficient $a$ of the function $j$.

The coefficient $a$ is the slope of the function $j^{-1}(y)$ yields $j$.

$x = \frac{y - 3}{2}$

$x = \frac{y}{2} + 3$

$y = \frac{x}{2} + 3$

$y = \frac{x - 3}{2}$