Function Analyser

Find a function's domain and range, test whether it's one-one, even or odd, find its inverse, and compose two functions.

f(x) =	(2x + 3)/(x - 1)
g(x) = (optional, for f∘g and g∘f)	x^2
Look from x =	-10
to x =	10

f(x) =

Use x, + − * / ^, brackets, sqrt, ln, sin, cos, exp, abs, π and e.

g(x) = (optional, for f∘g and g∘f)

Look from x =

to x =

Domain and range are worked out on this window; ℝ means no restriction at all.

f(x) = (2x + 3) / (x − 1)

Domain ℝ − {1}

Range: ℝ − {2}. f is one-one (injective). Its inverse is f⁻¹(x) = (3 + x) / (x − 2).

Exact form. Not independently checked.

Method: Restrictions from the expression, and the horizontal-line test
Worked out with: AllCalcHub
Check: Every step that can be checked was checked

Domain: ℝ − {1}
Range: ℝ − {2}
One-one?: Yes
Even or odd?: Neither
Inverse f⁻¹(x): (3 + x) / (x − 2)
f∘g (x): (2x^2 + 3) / (x^2 − 1)
g∘f (x): ((2x + 3) / (x − 1))^2

f, its inverse, and the mirror line y = x

Point at the curvePoint at the chart, or Tab to it and use ← →

f(x)f⁻¹(x)y = x

About the Function Analyser

The domain is every x the formula can take: no zero denominators, no square roots of negatives, no logarithms of zero or negatives. The range is every value it produces. A function has an inverse exactly when it's one-one: every output comes from only one input (the horizontal-line test).

For f(x) = (ax + b)/(cx + d), solve y = f(x) for x to get the inverse; the domain leaves out x = −d/c and the range leaves out y = a/c.

The formula

(f∘g)(x) = f(g(x)); f⁻¹(f(x)) = x

Worked example

f(x) = (2x + 3)/(x − 1)

Domain ℝ − {1}, range ℝ − {2}
y(x − 1) = 2x + 3 ⟹ x = (y + 3)/(y − 2)
f⁻¹(x) = (x + 3)/(x − 2)

Questions people ask

Why is the range approximate for some functions?

For anything beyond linear, quadratic and (ax + b)/(cx + d) functions, the range is read off the graph on the window you choose, and labelled approximate.

Is f∘g the same as g∘f?

Usually not: with f(x) = x + 1 and g(x) = x², f(g(x)) = x² + 1 but g(f(x)) = (x + 1)².