Here is a function. Function A: $x \to x + 3 \to y$ (a) Complete the table of values for function A. | x | y | |----|-----| | -5 | 11 | Here is another function. Function B: $x \to x + 5 \to y$ (b) Find the inverse function of function B. (c) Here is a composite function. $m \to \text{Function A} \to \text{Function B} \to 2p + 4$ Find an expression for m in terms of p. Give your answer in its simplest form.

Photo AI

Here is a function - OCR - GCSE Maths - Question 14 - 2017 - Paper 1

Question 14

Here is a function. Function A: $x \to x + 3 \to y$ (a) Complete the table of values for function A. | x | y | |----|-----| | -5 | 11 | Here is another fun... show full transcript

Worked Solution & Example Answer:Here is a function - OCR - GCSE Maths - Question 14 - 2017 - Paper 1

Step 1

Complete the table of values for function A.

96%

114 rated

Answer

To complete the table for function A, we start with the formula: $y = x + 3$ .

For $x = -5$ :
$y = -5 + 3 = -2$
Therefore, the first entry is $(-5, -2)$ .
For $y = 11$ :
Set the equation to $11 = x + 3$ .
Subtracting 3 from both sides gives:
$x = 11 - 3 = 8$ .
Thus, the second entry is $(8, 11)$ .

The completed table looks like this:

x	y
-5	-2
8	11

Step 2

Find the inverse function of function B.

99%

104 rated

Answer

The function B is given by:
$y = x + 5$ .
To find the inverse, we swap x and y:
$x = y + 5$ .
Solving for y, we have:
$y = x - 5$ . Thus, the inverse function is:
$f^{-1}(x) = x - 5$ .

Step 3

Find an expression for m in terms of p.

96%

101 rated

Answer

To find the expression for m, we need to evaluate the composite function.
Starting with function A:
$f_A(p) = p + 3$ . Next, substituting into function B:
$f_B(p + 3) = (p + 3) + 5 = p + 8$ . Finally, substituting into the expression $2p + 4$ :
We find:
$m = 2(p + 8) + 4 = 2p + 16 + 4 = 2p + 20$ . Hence, the final expression for m is:
$m = 2p + 20$ .

Join the GCSE students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;