Calculate \(f (1)\).

Find \(f ′(x)\).

Use your answer to part (b) to show that the x-coordinate of the local minimum point of the graph of \(f\) is 3.7 correct to 2 significant figures.

Find the range of \(f\).

Points A and B lie on the graph of \(f\). The x-coordinates of A and B are 1 and 7 respectively.

Write down the y-coordinate of B.

Points A and B lie on the graph of f . The x-coordinates of A and B are 1 and 7 respectively.

Find the gradient of the straight line passing through A and B.

M is the midpoint of the line segment AB.

Write down the coordinates of M.

L is the tangent to the graph of the function \(y = f (x)\), at the point on the graph with the same x-coordinate as M.

Find the gradient of L.

Find the equation of L. Give your answer in the form \(y = mx + c\).

\(\frac{{14}}{{(1)}} + (1) - 6\)     (M1)

Note: Award (M1) for substituting \(x = 1\) into \(f\).

\(= 9\)     (A1)(G2)

\( - \frac{{14}}{{{x^2}}} + 1\)     (A3)

Note: Award (A1) for \(-14\), (A1) for \(\frac{{14}}{{{x^2}}}\) or for \(x^{-2}\), (A1) for \(1\).

   Award at most (A2) if any extra terms are present.

\( - \frac{{14}}{{{x^2}}} + 1 = 0\) or \(f ' (x) = 0 \)     (M1)

Note: Award (M1) for equating their derivative in part (b) to 0.

\(\frac{{14}}{{{x^2}}} = 1\) or \({x^2} = 14\) or equivalent     (M1)

Note: Award (M1) for correct rearrangement of their equation.

\(x = 3.74165...(\sqrt {14})\)     (A1)

\(x = 3.7\)     (AG)

Notes: Both the unrounded and rounded answers must be seen to award the (A1). This is a “show that” question; appeals to their GDC are not accepted –award a maximum of (M1)(M0)(A0).

Specifically, \( - \frac{{14}}{{{x^2}}} + 1 = 0\) followed by \(x = 3.74165..., x = 3.7\) is awarded (M1)(M0)(A0).

\(1.48 \leqslant y \leqslant 9\)     (A1)(A1)(ft)(A1)

Note: Accept alternative notations, for example [1.48,9]. (\(x = \sqrt{14}\) leads to answer 1.48331...)

Note: Award (A1) for 1.48331…seen, accept 1.48378… from using the given answer \(x = 3.7\), (A1)(ft) for their 9 from part (a) seen, (A1) for the correct notation for their interval (accept \( \leqslant y \leqslant \) or \( \leqslant f \leqslant \) ).

3     (A1)

Note: Do not accept a coordinate pair.

\(\frac{{3 - 9}}{{7 - 1}}\)     (M1)

Note: Award (M1) for their correct substitution into the gradient formula.

\(= -1\)     (A1)(ft)(G2)

Note: Follow through from their answers to parts (a) and (e).

(4, 6)     (A1)(ft)(A1)

Note: Accept \(x = 4\), \(y = 6\). Award at most (A1)(A0) if parentheses not seen.

If coordinates reversed award (A0)(A1)(ft).

Follow through from their answers to parts (a) and (e).

\( - \frac{{14}}{{{4^2}}} + 1\)     (M1)

Note: Award (M1) for substitution into their gradient function.

Follow through from their answers to parts (b) and (g).

\( = \frac{1}{8}(0.125)\)     (A1)(ft)(G2)

\(y - 1.5 = \frac{1}{8}(x - 4)\)     (M1)(ft)(M1)

Note: Award (M1) for substituting their (4, 1.5) in any straight line formula,

(M1) for substituting their gradient in any straight line formula.

\(y = \frac{x}{8} + 4\)     (A1)(ft)(G2)

Note: The form of the line has been specified in the question.

Most candidates were able to evaluate the function and find the derivative for \(x + 6\) but the term with the negative index was problematic. The few candidates who equated their derivative to zero at the local minimum point progressed well and showed a thorough understanding of the differential calculus. Many did not attain full marks for the range of the function, either confusing this with the statistical concept of range or using the y-coordinate at B. Most were able to find the gradient and midpoint of the straight line passing through A and B. The final parts were also challenging for the majority: many had difficulty finding the gradient of the tangent L, instead using the slope formula for a straight line; the most common error in part (i) was to substitute in the coordinates of midpoint M rather than the point on the curve. Greater insight into the problem would have come from using the given sketch of the curve and annotating it; it seems that many candidates do not link the algebraic nature of the differential calculus with the curve in question.

Most candidates were able to evaluate the function and find the derivative for \(x + 6\) but the term with the negative index was problematic. The few candidates who equated their derivative to zero at the local minimum point progressed well and showed a thorough understanding of the differential calculus. Many did not attain full marks for the range of the function, either confusing this with the statistical concept of range or using the y-coordinate at B. Most were able to find the gradient and midpoint of the straight line passing through A and B. The final parts were also challenging for the majority: many had difficulty finding the gradient of the tangent L, instead using the slope formula for a straight line; the most common error in part (i) was to substitute in the coordinates of midpoint M rather than the point on the curve. Greater insight into the problem would have come from using the given sketch of the curve and annotating it; it seems that many candidates do not link the algebraic nature of the differential calculus with the curve in question.

Most candidates were able to evaluate the function and find the derivative for \(x + 6\) but the term with the negative index was problematic. The few candidates who equated their derivative to zero at the local minimum point progressed well and showed a thorough understanding of the differential calculus. Many did not attain full marks for the range of the function, either confusing this with the statistical concept of range or using the y-coordinate at B. Most were able to find the gradient and midpoint of the straight line passing through A and B. The final parts were also challenging for the majority: many had difficulty finding the gradient of the tangent L, instead using the slope formula for a straight line; the most common error in part (i) was to substitute in the coordinates of midpoint M rather than the point on the curve. Greater insight into the problem would have come from using the given sketch of the curve and annotating it; it seems that many candidates do not link the algebraic nature of the differential calculus with the curve in question.

Most candidates were able to evaluate the function and find the derivative for \(x + 6\) but the term with the negative index was problematic. The few candidates who equated their derivative to zero at the local minimum point progressed well and showed a thorough understanding of the differential calculus. Many did not attain full marks for the range of the function, either confusing this with the statistical concept of range or using the y-coordinate at B. Most were able to find the gradient and midpoint of the straight line passing through A and B. The final parts were also challenging for the majority: many had difficulty finding the gradient of the tangent L, instead using the slope formula for a straight line; the most common error in part (i) was to substitute in the coordinates of midpoint M rather than the point on the curve. Greater insight into the problem would have come from using the given sketch of the curve and annotating it; it seems that many candidates do not link the algebraic nature of the differential calculus with the curve in question.

Most candidates were able to evaluate the function and find the derivative for \(x + 6\) but the term with the negative index was problematic. The few candidates who equated their derivative to zero at the local minimum point progressed well and showed a thorough understanding of the differential calculus. Many did not attain full marks for the range of the function, either confusing this with the statistical concept of range or using the y-coordinate at B. Most were able to find the gradient and midpoint of the straight line passing through A and B. The final parts were also challenging for the majority: many had difficulty finding the gradient of the tangent L, instead using the slope formula for a straight line; the most common error in part (i) was to substitute in the coordinates of midpoint M rather than the point on the curve. Greater insight into the problem would have come from using the given sketch of the curve and annotating it; it seems that many candidates do not link the algebraic nature of the differential calculus with the curve in question.

Most candidates were able to evaluate the function and find the derivative for \(x + 6\) but the term with the negative index was problematic. The few candidates who equated their derivative to zero at the local minimum point progressed well and showed a thorough understanding of the differential calculus. Many did not attain full marks for the range of the function, either confusing this with the statistical concept of range or using the y-coordinate at B. Most were able to find the gradient and midpoint of the straight line passing through A and B. The final parts were also challenging for the majority: many had difficulty finding the gradient of the tangent L, instead using the slope formula for a straight line; the most common error in part (i) was to substitute in the coordinates of midpoint M rather than the point on the curve. Greater insight into the problem would have come from using the given sketch of the curve and annotating it; it seems that many candidates do not link the algebraic nature of the differential calculus with the curve in question.

Most candidates were able to evaluate the function and find the derivative for \(x + 6\) but the term with the negative index was problematic. The few candidates who equated their derivative to zero at the local minimum point progressed well and showed a thorough understanding of the differential calculus. Many did not attain full marks for the range of the function, either confusing this with the statistical concept of range or using the y-coordinate at B. Most were able to find the gradient and midpoint of the straight line passing through A and B. The final parts were also challenging for the majority: many had difficulty finding the gradient of the tangent L, instead using the slope formula for a straight line; the most common error in part (i) was to substitute in the coordinates of midpoint M rather than the point on the curve. Greater insight into the problem would have come from using the given sketch of the curve and annotating it; it seems that many candidates do not link the algebraic nature of the differential calculus with the curve in question.

Most candidates were able to evaluate the function and find the derivative for \(x + 6\) but the term with the negative index was problematic. The few candidates who equated their derivative to zero at the local minimum point progressed well and showed a thorough understanding of the differential calculus. Many did not attain full marks for the range of the function, either confusing this with the statistical concept of range or using the y-coordinate at B. Most were able to find the gradient and midpoint of the straight line passing through A and B. The final parts were also challenging for the majority: many had difficulty finding the gradient of the tangent L, instead using the slope formula for a straight line; the most common error in part (i) was to substitute in the coordinates of midpoint M rather than the point on the curve. Greater insight into the problem would have come from using the given sketch of the curve and annotating it; it seems that many candidates do not link the algebraic nature of the differential calculus with the curve in question.

Most candidates were able to evaluate the function and find the derivative for \(x + 6\) but the term with the negative index was problematic. The few candidates who equated their derivative to zero at the local minimum point progressed well and showed a thorough understanding of the differential calculus. Many did not attain full marks for the range of the function, either confusing this with the statistical concept of range or using the y-coordinate at B. Most were able to find the gradient and midpoint of the straight line passing through A and B. The final parts were also challenging for the majority: many had difficulty finding the gradient of the tangent L, instead using the slope formula for a straight line; the most common error in part (i) was to substitute in the coordinates of midpoint M rather than the point on the curve. Greater insight into the problem would have come from using the given sketch of the curve and annotating it; it seems that many candidates do not link the algebraic nature of the differential calculus with the curve in question.