The graphical relationship between f and its inverse
Functions
The graphical relationship between f and its inverse
4 self-marking Additional Maths practice questions on The graphical relationship between f and its inverse, written specifically for Cambridge IGCSE Additional Maths (0606) across Single tier. Instant marking with full worked solutions on every question.
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A few of the 4 questions, from easier to harder. Choose your exam board and tier to practise them, with every answer marked.
Question 12 marksGrade 7
The point (2,7) lies on the graph of y=f(x). Write down the coordinates of the corresponding point on the graph of y=f−1(x)
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Fails to swap the x and y coordinates of a point when finding its image under a reflection or inverse relationship that requires it
The graph of y=f−1(x) is the reflection of y=f(x)
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(a) State the
x
-coordinate. [1]
(b) State the
y
-coordinate. [1]
Show the worked solution
Reflecting a point in the line y=x swaps its coordinates.
So (2,7) on y=f(x) corresponds to (7,2) on y=f−1(x).
Question 22 marksGrade 8
The function f is defined for x∈R by f(x)=2x+3. The graphs of y=f(x) and y=f−1(x) intersect the line y=x at the same point.
Find the coordinates of this point. [2]
Show the worked solution
Since y=f(x) and y=f−1(x) are reflections of each other in the line y=x
Question 32 marksGrade 8
The curve y=f(x) crosses the y-axis at (0,5) and the x-axis at (2,0).
State the coordinates of the two points where the curve y=f−1(x) crosses the axes. [2]
Question 42 marksGrade 9
The function g has inverse g−1(x)=3x−2. The graphs of y=g(x) and y=g−1(x) intersect the line y=x at the same point.
Find the coordinates of this point. [2]
Leaves coordinates unchanged under an inverse-function reflection instead of swapping the x- and y-coordinates
The graph of y=f−1(x) is the REFLECTION of y=f(x) in the line y=x, which swaps the x- and y-coordinate of every point - (0,5) on f becomes (5,0) on f−1, and (2,0) becomes (0,2), not the same points unchanged.
Sets up the wrong equation to solve, using zero instead of the condition actually specified by the problem
Since f and f−1 are reflections of each other in the line y=x, they meet y=x at exactly the point where f(x)=x - solve 2x+3=x, not 2x+3=0.