Find \(f'(x)\) .

Find the x-coordinate of M.

Find the x-coordinate of N.

The line L is the tangent to the curve of f at \((3{\text{, }}12)\). Find the equation of L in the form \(y = ax + b\) .

\(f'(x) = {x^2} + 4x - 5\)     A1A1A1     N3

[3 marks]

evidence of attempting to solve \(f'(x) = 0\)     (M1)

evidence of correct working     A1

e.g. \((x + 5)(x - 1)\) , \(\frac{{ - 4 \pm \sqrt {16 + 20} }}{2}\) , sketch

\(x = - 5\), \(x = 1\)     (A1)

so \(x = - 5\)     A1     N2

[4 marks]

METHOD 1

\(f''(x) = 2x + 4\) (may be seen later)     A1

evidence of setting second derivative = 0     (M1)

e.g. \(2x + 4 = 0\)

\(x = - 2\)     A1     N2

METHOD 2

evidence of use of symmetry     (M1)

e.g. midpoint of max/min, reference to shape of cubic

correct calculation     A1

e.g. \(\frac{{ - 5 + 1}}{2}\)

\(x = - 2\)     A1     N2

[3 marks]

attempting to find the value of the derivative when \(x = 3\)     (M1)

\(f'(3) = 16\)     A1

valid approach to finding the equation of a line     M1

e.g. \(y - 12 = 16(x - 3)\) , \(12 = 16 \times 3 + b\)

\(y = 16x - 36\)     A1     N2

[4 marks]

This question was very well done with most candidates showing their work in an orderly manner. There were a number of candidates, however, who were a bit sloppy in indicating when a function was being equated to zero and they “solved” an expression rather than an equation.

This question was very well done with most candidates showing their work in an orderly manner. There were a number of candidates, however, who were a bit sloppy in indicating when a function was being equated to zero and they “solved” an expression rather than an equation. Many candidates went through first and second derivative tests to verify that the point they found was a maximum or an inflexion point; this was unnecessary since the graph was given. Many also found the y-coordinate which was unnecessary and used up valuable time on the exam.

This question was very well done with most candidates showing their work in an orderly manner. There were a number of candidates, however, who were a bit sloppy in indicating when a function was being equated to zero and they “solved” an expression rather than an equation. Many candidates went through first and second derivative tests to verify that the point they found was a maximum or an inflexion point; this was unnecessary since the graph was given. Many also found the y-coordinate which was unnecessary and used up valuable time on the exam.

This question was very well done with most candidates showing their work in an orderly manner. There were a number of candidates, however, who were a bit sloppy in indicating when a function was being equated to zero and they “solved” an expression rather than an equation. Many candidates went through first and second derivative tests to verify that the point they found was a maximum or an inflexion point; this was unnecessary since the graph was given. Many also found the y-coordinate which was unnecessary and used up valuable time on the exam.