Combination of Functions
When adding, subtracting and multipiying, the domain of the new function will contain the x-values common to the domains of both original functions.
When dividing functions the domain is further restricted such that the denoinator does not equal zero.
COMBINATION OF FUNCTIONS
Let f and g be two functions. Then for values of x in the domain of both f and g the sum, difference, product and quotient of the two functions are defined as follows:
º Difference:$\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)$
º Product: $\left(f\xb7g\right)\left(x\right)=f\left(x\right)\xb7g\left(x\right)$
º Quotient: $\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$, $g\left(x\right)\ne 0$
Let's look at a couple examples.
Step 1: Find the sum of the two functions. |
$\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$ Sum Rule $\left(3x-2\right)+\left({x}^{2}+5x-7\right)$ Substitute Combine like terms: $\left(f+g\right)\left(x\right)={x}^{2}+8x-9$ |
Step 2: Find the difference of the two functions. |
$\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)$ Diff. Rule $\left(3x-2\right)-\left({x}^{2}+5x-7\right)$ Substitute Distribute the negative: $3x-2-{x}^{2}-5x+7$ Combine like terms: $\left(f-g\right)\left(x\right)=-{x}^{2}-2x+5$ |
Step 3: Find the domain of the new functions. Find the domain of f(x) and the domain of g(x) and determine the intersection of these domains. This intersection is the domain of the sum and difference of the functions. |
The domain of f(x) = 3x - 2 is the set of all real numbers. The domain of g(x) = x^{2} + 5x - 7 is the set of all real numbers. Therefore the domain of (f + g)(x) and (f - g)(x) is the set of all real numbers. |
Step 4: Evaluate the new functions with the given value of x. |
${\left(}f+g{\right)}\left(1\right)={1}^{2}+8\left(1\right)-9={0}$ $\left(f-g\right)\left(1\right)=-{\left(1\right)}^{2}-2\left(1\right)+5={2}$ |
Step 1: Find the product of the two functions. |
$\left(f\xb7g\right)\left(x\right)=f\left(x\right)\xb7g\left(x\right)$ $\left(4{x}^{2}\right)\xb7\left(5x-2\right)$ $\left(f\xb7g\right)\left(x\right)=20{x}^{2}-8{x}^{2}$ |
Step 2: Find the quotient of the two functions. |
$\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$ $\left(\frac{f}{g}\right)\left(x\right)=\frac{4{x}^{2}}{5x-2}$ |
Step 3: Find the domain of the new functions. Find the domain of f(x) and the domain of g(x) and determine the intersection of these domains. This intersection is the domain of the product of the functions. The domain of the quotient is further restricted such that g(x) = 0. |
The domain of f(x) = 4x^{2} is the set of all real numbers. The domain of g(x) = 5x - 2 is the set of all real numbers. Therefore the domain of $\left(f\xb7g\right)\left(x\right)$ is the set of all real numbers. However the domain of $\left(\frac{f}{g}\right)\left(x\right)$ cannot include the x-value when g(x) = 0. $0=5x-2\to 2=5x\to \frac{2}{5}=x$ Therefore the domain of $\left(\frac{f}{g}\right)\left(x\right)$ is all real numbers except $\frac{2}{5}$. |
Step 4: Evaluate the new functions with the given value of x. |
$\left(f\xb7g\right)\left(2\right)=20{x}^{2}-8{x}^{2}=4{\left(2\right)}^{2}={48}$ $\left(\frac{f}{g}\right)\left(2\right)=\frac{4{x}^{2}}{5x-2}=\frac{16}{8}={2}$ |
Related Links: Math algebra Composition of Functions Extreme Value Theorem Calculus Topics |
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