21N.1.SL.TZ0.5

pestleMathematicsAASLPaper 121N· sl-2-1-equations-of-straight-lines-parallel-and-perpendicularsource ↗

The function $f$ is defined for all $x \in ℝ$ . The line with equation $y = 6 x - 1$ is the tangent to the graph of $f$ at $x = 4$ .

The function $g$ is defined for all $x \in ℝ$ where $g (x) = x^{2} - 3 x$ and $h (x) = f (g (x))$ .

Write down the value of $f' (4)$ .

[1]

Find $f (4)$ .

[1]

Find $h (4)$ .

[2]

Hence find the equation of the tangent to the graph of $h$ at $x = 4$ .

[3]

Markscheme / solution

$f' (4) = 6$ A1

[1 mark]

$f (4) = 6 \times 4 - 1 = 23$ A1

[1 mark]

$h (4) = f (g (4))$ (M1)

$h (4) = f (4^{2} - 3 \times 4) = f (4)$

$h (4) = 23$ A1

[2 marks]

attempt to use chain rule to find $h'$ (M1)

$f' (g (x)) \times g' (x)$ OR $(x^{2} - 3 x)' \times f' (x^{2} - 3 x)$

$h' (4) = (2 \times 4 - 3) f' (4^{2} - 3 \times 4)$ A1

$= 30$

$y - 23 = 30 (x - 4)$ OR $y = 30 x - 97$ A1

[3 marks]

Examiners’ report

[N/A]