Function notation is how a function is represented algebraically. It can be expressed as \(f(x)\) where:
The way that \(f(x)\) is used to represent the function's output is similar to how we used \(y\) in previous years. For example, \(f(x) = 2x\) is equivalent to \(y = 2x\).
For the sake of simplicity, we will only focus on function notation that uses \(1\) input variable.
Function notation doesn't necessarily need to be in the form \(f(x)\). It can use different variables for both its function name and input. For example, \(g(t)\) or \(\cos(\theta)\).
\(A(l)\)
We can identify the following characteristics as such:
\(\sin^{-1}(0.5)\)
We can identify the following characteristics as such:
\(\sqrt{9}\)
We can identify the following characteristics as such:
\((2x)²\)
We can identify the following characteristics as such:
When given a function and input value, we substitute the input variable for the value and then simplify if necessary. We can follow this method in order to find the result of \(f(x)=2x+5\) when \(x=4\).
All we need to do is substitute \(4\) for \(x\) and simplify to find the final result.
\(f(x)=2x+5\)
\(f(4)=2(4)+5\)
\(f(4)=8+5\)
\(f(4)=13\)
Therefore, we can determine the result of \(f(x)=2x+5\) when \(x=4\) is \(\boldsymbol{13}\).
Determine the outputs given the following functions:
\(f(x) = 2x⁴ - 3x²\)
\(h(a)= \sqrt{a-5}\)
\(g(k)= \cfrac{6k}{k+1}\)
\(h(69)\)
We can substitute \(69\) for \(a\) since \(69\) represents the input while \(a\) represents the input variable. We can then simplify the expression to get the result:
\(h(a) = \sqrt{a-5}\)
\(h(69) = \sqrt{69-5}\)
\(h(69) = \sqrt{64}\)
\(h(69) = 8\)
Therefore, we can determine that \(\boldsymbol{h(69) = 8}\).
\(f(-3)+4g(3)\)
First, we can substitute \(-3\) for \(x\) since \(-3\) represents the input while \(x\) represents the input variable:
\(f(x) = 2x⁴ - 3x²\)
\(f(-3) = 2(-3)⁴ - 3(-3)²\)
\(f(-3) = 2(81) - 3(9)\)
\(f(-3) = 162 - 27\)
\(f(-3) = 135\)
Next, we can substitute \(3\) for \(k\) since \(3\) represents the input while \(x\) represents the input variable. We can then multiply the result by \(4\):
\(g(k)= \cfrac{6k}{k+1}\)
\(4g(3)= 4\left(\cfrac{6(3)}{3+1}\right)\)
\(4g(3)= \cancel4\left(\cfrac{18}{\cancel4}\right)\)
\(4g(3)= 18\)
Finally, we can add the results of both expressions to get the final result:
\(f(-3)+4g(3) = 135 + 18\)
\(f(-3)+4g(3) = 153\)
Therefore, we can determine that \(\boldsymbol{f(-3)+4g(3) = 153}\).
\(2g\left(\cfrac{1}{2}\right)f(-1)\)
First, we can substitute \(\cfrac{1}{2}\) for \(k\) since \(\cfrac{1}{2}\) represents the input while \(k\) represents the input variable. We can then multiply the result by \(2\):
Next, we can substitute \(-1\) for \(x\) since \(-1\) represents the input while \(x\) represents the input variable:
\(f(x) = 2x⁴ - 3x²\)
\(f(-1) = 2(-1)⁴ - 3(-1)²\)
\(f(-1) = 2(1) - 3(1)\)
\(f(-1) = 2 - 3\)
\(f(-1) = -1\)
Finally, we can multiply the results of both expressions to get the final result:
\(2g\left(\cfrac{1}{2}\right)f(-1) = (4)(-1)\)
\(2g\left(\cfrac{1}{2}\right)f(-1) = -4\)
Therefore, we can determine that \(\boldsymbol{2g\left(\cfrac{1}{2}\right)f(-1) = -4}\).
\(h(\sin(\theta))\)
First, we can substitute \(\sin(\theta)\) for \(a\) since \(\sin(\theta)\) represents the input while \(a\) represents the input variable:
\(h(a) = \sqrt{a-5}\)
\(h(\sin(\theta)) = \sqrt{\sin(\theta) - 5}\)
Therefore, we can determine that \(\boldsymbol{h(\sin(\theta)) = \sqrt{\sin(\theta) - 5}}\).