Function Notation
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Question 1 of 7
1. Question
Given the function `f(x)=x+4`, solve for:`(i) f(1)``(ii) f(-3)``(iii) f(0)`-
`(i) f(1)=` (5)`(ii) f(-3)=` (1)`(iii) f(0)=` (4)
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Function notation is a brief shortcut way to give information about a functionSimply substitute the given values as `x``f(1)``f(x)` `=` `x+4` `f(1)` `=` `1+4` Substitute `x=1` `=` `5` `f(-3)``f(x)` `=` `x+4` `f(-3)` `=` `(-3)+4` Substitute `x=-3` `=` `1` `f(0)``f(x)` `=` `x+4` `f(0)` `=` `0+4` Substitute `x=0` `=` `4` `f(1)=5``f(-3)=1``f(0)=4` -
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Question 2 of 7
2. Question
Given the function `f(x)=2x^3-4x+3`, solve for:`(i) f(-3)``(ii) f(2)`-
`(i) f(-3)=` (-39)`(ii) f(2)=` (11)
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Function notation is a brief shortcut way to give information about a functionSimply substitute the given values as `x``f(-3)``f(x)` `=` `2x^3-4x+3` `f(-3)` `=` `[2*(-3)^3]-[4*(-3)]+3` Substitute `x=-3` `=` `[2*(-27)]+12+3` `=` `-54+15` `=` `-39` `f(2)``f(x)` `=` `2x^3-4x+3` `f(2)` `=` `[2*(2)^3]-[4(2)+3]` Substitute `x=2` `f(2)` `=` `(2*8)-8+3` `f(2)` `=` `16-5` `=` `11` `f(-3)=-39``f(2)=11` -
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Question 3 of 7
3. Question
Solve for `f(-1)-f(3)`, given that:$$f(x)=\begin{cases}x^3-2, x≥2\\2x^2+4x-1,x<2\end{cases}$$- `f(-1)-f(3)=` (-28)
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Function notation is a brief shortcut way to give information about a functionPiecewise Functions (aka Composite Functions) are a set of functions that share a common pointFirst, substitute the given values as `x` to their respective functions`f(x)` `=` $$\begin{cases}x^3-2, x≥2\\2x^2+4x-1,x<2\end{cases}$$ `f(-1)``f(x)` `=` `2x^2+4x-1` Second function `f(-1)` `=` `[2*(-1)^2]+[4*(-1)]-1` Substitute `x=-1` `=` `2-4-1` Evaluate `=` `-3` `f(3)``f(x)` `=` `x^3-2` First function `f(3)` `=` `(3)^3-2` Substitute `x=3` `=` `27-2` Evaluate `=` `25` Finally, substitute new values of the functions and solve for the final value$$\color{#00880A}{f(-1)}-\color{#9a00c7}{f(3)}$$ `=` $$\color{#00880A}{-3}-\color{#9a00c7}{25}$$ Substitute known values `=` `-28` Evaluate `-28` -
Question 4 of 7
4. Question
Solve for `f(4)-f(0)+f(-2)`, given that:$$f(x)=\begin{cases}3x, x≥3\\1,-1<x<3\\x^3-2,x≤-1\end{cases}$$- `f(4)-f(0)+f(-2)=` (1)
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Function notation is a brief shortcut way to give information about a functionPiecewise Functions (aka Composite Functions) are a set of functions that share a common pointFirst, substitute the given values as `x` to their respective functions`f(x)` `=` $$\begin{cases}3x, x≥3\\1,-1<x<3\\x^3-2,x≤-1\end{cases}$$ `f(4)``f(x)` `=` `3x` First function `f(4)` `=` `3(4)` Substitute `x=4` `=` `12` Evaluate `f(0)``f(x)` `=` `1` Second function `f(0)` `=` `1` Substitute `x=0` `f(0)``f(x)` `=` `x^3-2` Third function `f(-2)` `=` `(-2)^3-2` Substitute `x=-2` `=` `-8-2` Evaluate `=` `-10` Finally, substitute new values of the functions and solve for the final value$$\color{#00880A}{f(4)}-\color{#9a00c7}{f(0)}+\color{#e65021}{f(-2)}$$ `=` $$\color{#00880A}{12}-\color{#9a00c7}{1}+(\color{#e65021}{-10})$$ Substitute known values `=` `1` Evaluate `1` -
Question 5 of 7
5. Question
Solve for `f(2)-f(-2)+f(-1)`, given that:$$f(x)=\begin{cases}2x-4, x≥1\\x+1,-1<x<1\\x^2,x≤-1\end{cases}$$- `f(2)-f(-2)+f(-1)=` (-3)
Hint
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Function notation is a brief shortcut way to give information about a functionPiecewise Functions (aka Composite Functions) are a set of functions that share a common pointFirst, substitute the given values as `x` to their respective functions`f(x)` `=` $$\begin{cases}2x-4, x≥1\\x+1,-1<x<1\\x^2,x≤-1\end{cases}$$ `f(4)``f(x)` `=` `2x-4` First function `f(2)` `=` `2(2)-4` Substitute `x=2` `=` `4-4` Evaluate `=` `0` `f(-2)``f(x)` `=` `x^2` Third function `f(-2)` `=` `(-2)^2` Substitute `x=-2` `=` `4` Evaluate `f(-1)``f(x)` `=` `x^2` Third function `f(-1)` `=` `(-1)^2` Substitute `x=-1` `=` `1` Evaluate Finally, substitute new values of the functions and solve for the final value$$\color{#00880A}{f(2)}-\color{#9a00c7}{f(-2)}+\color{#e65021}{f(-1)}$$ `=` $$\color{#00880A}{0}-\color{#9a00c7}{4}+\color{#e65021}{1}$$ Substitute known values `=` `-3` Evaluate `-3` -
Question 6 of 7
6. Question
Solve for `g(-4)+g(-2)+g(3)`, given that:$$g(x)=\begin{cases}0, x≤1\\-5,-3<x<0\\2x,x≥0\end{cases}$$- `g(-4)+g(-2)+g(3)=` (1)
Hint
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Correct!
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Function notation is a brief shortcut way to give information about a functionPiecewise Functions (aka Composite Functions) are a set of functions that share a common pointFirst, substitute the given values as `x` to their respective functions`g(x)` `=` $$\begin{cases}0, x≤1\\-5,-3<x<0\\2x,x≥0\end{cases}$$ `g(-4)``g(x)` `=` `0` First function `g(-4)` `=` `0` Substitute `x=-4` `g(-2)``g(x)` `=` `-5` Second function `g(-2)` `=` `-5` Substitute `x=-2` `g(3)``g(x)` `=` `2x` Third function `g(3)` `=` `2(3)` Substitute `x=3` `=` `6` Evaluate Finally, substitute new values of the functions and solve for the final value$$\color{#00880A}{g(-4)}+\color{#9a00c7}{g(-2)}+\color{#e65021}{g(3)}$$ `=` $$\color{#00880A}{0}+(\color{#9a00c7}{-5})+\color{#e65021}{6}$$ Substitute known values `=` `1` Evaluate `1` -
Question 7 of 7
7. Question
Given the function `f(x)=x^2+3x+5`, solve$$\frac{f(2+h)-f(2)}{h}, h≠0$$- (7+h)
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Function notation is a brief shortcut way to give information about a functionFirst, substitute the given values as `x` to the given function`f(2+h)``f(x)` `=` `x^2+3x+5` `f(2+h)` `=` `(2+h)^2+3(2+h)+5` Substitute `x=2+h` `=` `4+4h+h^2+6+3h+5` Evaluate `=` `15+7h+h^2` Combine like terms `f(2)``f(x)` `=` `x^2+3x+5` `f(2+h)` `=` `(2)^2+3(2)+5` Substitute `x=2` `=` `4+6+5` Evaluate `=` `15` Finally, substitute new values of the functions and solve for the final value$$\frac{\color{#00880A}{f(2+h)}-\color{#9a00c7}{f(2)}}{h}$$ `=` $$\frac{\color{#00880A}{15+7h+h^2}-\color{#9a00c7}{15}}{h}$$ Substitute known values `=` `(7h+h^2)/h` Evaluate `=` `7+h` Simplify `(f(2+h)-f(2))/h=7+h`