Using a Given (Algebraic) Rule
Representing Functions with an Algebraic Rule
No matter how a function is defined, a function is a rule that takes an input, does something with it, then produces an output.
We can define a function using an algebraic rule, which describes what the function (or “machine”) does to the input in order to produce an output.
For example, the function \(f(x)=x^2+5\) describes what happens when you plug in an arbitrary \(x\)-value. That is, \(x\) serves a “place-holder” or “representative” for a number, and the rule tells you what operations would be performed if \(x\) were replaced with an actual value. So, in the case of \(f\) above, \(f\) takes some input \(x\), squares it, then adds \(5\) to the result. So, if \(x=3\), then \(f(3)=3^2+5=14\). The output of \(f\) on input \(x=3\), in this case, is \(14\).
Often, the outputs of a function are referred to as \(y\)-values, and the inputs are referred to as \(x\)-values. This allows us to graph functions on an \(xy\)-coordinate graph (in the same way you’d go about graphing a line or a parabola, by plugging in \(x\)-values and plotting the corresponding \(y\)-value above the given \(x\)-value.)
\(f(2)=2^2-10=-6\) and \(f(-2)=(-2)^2-10=-6\).
Every place you see an \(x\) in the function rule, replace it with the given input (i.e. the number in parentheses next to the \(f\).
\(g(1)=3(1)^2-5(1)+1=0\) and \(g(-3)=3(-3)^2-5\cdot (-3)+2=44\).
Every place you see an \(x\) in the function rule, replace it with the given input (i.e. the number in parentheses next to the \(g\).
Solution: The problem is basically asking “what is \(h(1)\)”, “what is \(h(4)\)”, and “what is \(h(0)\)”. The inputs are the numbers inside the parentheses. The outputs are the numbers you get when you plug those inputs into the rule for \(h\).
\(h(1)=\frac{1+\sqrt{1}}{1}=\frac{2}{1}= 2\)
\(h(4)=\frac{1+\sqrt{4}}{4}=\frac{3}{4}\)
\(h(0)=\frac{1+\sqrt{0}}{0}\) which is not defined because you can’t have division by zero. Sometimes functions are not defined on certain inputs; that is, some inputs don’t have any output when applying the rule for the function. When this happens, we simply say that “the function is not defined on the given input”
In this case, we are given inputs \(2, a, \) and \(x+1\).
To compute \(f(2)\), one simply replaces all the \(x\)’s in the function’s rule with \(2\) and evaluates the result of \(2^3+5(2)^2+2\) to get the final answer of \(f(2)=30\).
This works the exact same way when you plug in variables or expressions into a function, because variables and expressions represent numbers!
To evaluate \(f(a)\) simply replace all your \(x\)’s in the function’s rule with \(a\). You will get that \(f(a)=a^3+5a^2+2\). The problem said “don’t simplify” so we can leave this as is.
Similarly, to evaluate \(f(x+1)\), simply replace all \(x\)’s in the function’s rule with the entire expression \(x+1\), just like you would with numbers. I recommend that you keep the expression \(x+1\) in parentheses when you plug the whole thing into your function.
Thus, \(f(x+1)=(x+1)^3+5(x+1)^2+2\). All we did was replace the \(x\)’s in the rule with \((x+1)\).
Solution: This problem works exactly the same way as the last problem.
For \(g\left(\frac{1}{x}\right)\), replace all \(x\)’s in the function rule with \(\frac{1}{x}\).
This gives you \(g\left(\frac{1}{x}\right)=\sqrt{4+\frac{1}{x}}\). This really can’t be simplified any more.
Using the same reasoning, \(g(x-1)=\sqrt{4+(x-1)}=\sqrt{4+x-1}=\sqrt{3+x}\), which can’t be simplified any more than that.