AI HL · 12.5
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Function notation, domain and range
12.5.1
Screenshot and video to come
The volume of water in a tank is \(V(t)=480-15t\) litres, where t is the time in minutes after a tap is opened. Find \(V(8)\), and find the value of t for which \(V(t)=105\).
Have a go first. Then open the working one line at a time.
\(V(8)\) means: put \(t=8\) into the formula
V is the name of the function and the number in the bracket is the input.
\(V(8)=480-15\times8=360\)
After 8 minutes there are 360 litres left.
\(480-15t=105 \;\Rightarrow\; 15t=375 \;\Rightarrow\; t=25\)
\(V(t)=105\) is an equation: the output is given and you find the input.
\(f(x)=\sqrt{x-2}+3\). Write down the domain and the range of f.
Have a go first. Then open the working one line at a time.
\(x-2\geq0 \;\Rightarrow\; x\geq2\)
You cannot take the square root of a negative number.
\(\sqrt{x-2}\geq0\), so \(f(x)\geq3\)
The square root is never negative. Adding 3 lifts the smallest output to 3.
Domain \(x\geq2\)\; range \(y\geq3\)
Write each as an inequality.
\(g(x)=-x^2+6x-4\), for all real x. Find the range of g.
Have a go first. Then open the working one line at a time.
The coefficient of \(x^2\) is negative, so the graph is a hill with a maximum.
A hill has no lowest point, so the range is bounded above only.
From the calculator: maximum at \((3,\,5)\)
Graph g and use the maximum tool.
\(y\leq5\)
Every output is 5 or less.
The graph shows \(y=f(x)\). The closed dot at \((-2,\,7)\) is part of the graph. The open dot at \((4,\,7)\) is not. Write down the domain and the range of f.
Have a go first. Then open the working one line at a time.
From \(x=-2\) (included) to \(x=4\) (left out): \(-2\leq x<4\)
A closed dot gives \(\leq\)\; an open dot gives \(<\).
The lowest point is the minimum at \((1,\,-2)\). The highest value is 7, reached at \(x=-2\).
The open dot also sits at 7, but 7 is still reached because the closed dot is on the graph.
Domain \(-2\leq x<4\)\; range \(-2\leq y\leq7\)
Range comes from the y-values.
A ball is thrown upwards from a height of 1.5 m. Its height after t seconds is \(h(t)=-5t^2+20t+1.5\) metres. State a reasonable domain for the model. Give the end point to 3 significant figures.
Have a go first. Then open the working one line at a time.
Time cannot be negative, and the throw starts at \(t=0\).
The model begins when the ball is thrown.
Solve \(h(t)=0\). From the calculator: \(t=4.07\)
The ball stops when it lands. Use the positive solution.
\(0\leq t\leq4.07\)
Outside this interval the height is negative or the ball has not been thrown.