Functions and Function Notation - GRE Quantitative Reasoning

$(g\circ f)(2)=10$. Evaluate (f(2)) first, then apply (g) to that output for the composition.

$(f\circ g)(2)=7$. First compute (g(2)), then input that result into (f) to find the composed value.

$(f\circ g)(x)=f(g(x))$. Function composition applies (f) to the result of (g(x)), nesting the functions.

$f^{-1}(x)$ is inverse; $\frac{1}{f(x)}$ is reciprocal. The inverse undoes the function's operation, while the reciprocal is the multiplicative inverse of the output.

$f^{-1}(y)=x$ exactly when $f(x)=y$. The inverse function reverses the mapping, so inputs and outputs are swapped.

$f(1)=2$. As (1 < 2), apply the first piece of the piecewise definition for computation.

$f(2)=6$. Since (2 geq 2), use the second piece of the piecewise function to evaluate.

$(-\infty,5]$. Ensure the radicand is non-negative for the square root function to be defined over reals.

$(-\infty,5)\cup(5,\infty)$. Exclude the value where the denominator is zero to ensure the function is defined for all real inputs.

$[-1,\infty)$. Require the expression inside the square root to be non-negative to yield real outputs.

$f(x-3)=(x-3)^2$. Substitute the expression (x-3) into the quadratic function in place of (x).

$f(a+1)=2a-5$. Replace (x) with (a+1) in the linear expression and simplify algebraically.

$h(0)=\frac{2}{3}$. Insert (x = 0) into the rational function and reduce the fraction.

$g(4)=15$. Evaluate the quadratic function by plugging in (x = 4) and simplifying.

$f(-2)=-11$. Substitute (x = -2) into the linear function to compute the output directly.

The set of all possible output values $f(x)$. The range includes every possible output obtained by applying the function to values in its domain.

The set of all allowed input values $x$. The domain consists of every input (x) for which the function is defined and produces a valid output.

Input $a$ maps to output $b$ under $f$. The equation states that applying function (f) to input (a) produces the output value (b).

$f^{-1}(9)=3$. With the domain restriction ensuring one-to-one, solve (x^2 = 9) and take the non-negative root.

$f(x)$ is the output value of function $f$ for input $x$. This notation evaluates the function (f) at the specific input value (x), yielding the corresponding output.

It implies $x=y$. One-to-one functions are injective, so equal outputs necessitate equal inputs.

$f(3)-f(1)=8$. Compute each function value separately using the quadratic formula, then subtract.

$g(f(x))=x^2-1$. Input (f(x)) into (g), resulting in subtraction after squaring.