Explain why the graph of $f$ has a local minimum when $x = 5$.

Find the set of values of $x$ for which the graph of $f$ is concave down.

The following diagram shows the shaded regions $A$, $B$ and $C$.

The regions are enclosed by the graph of $f'$, the $x$-axis, the $y$-axis, and the line $x = 6$.

The area of region $A$ is 12, the area of region $B$ is 6.75 and the area of region $C$ is 6.75.

Given that $f(0) = 14$, find $f(6)$.

The following diagram shows the shaded regions $A$, $B$ and $C$.

The regions are enclosed by the graph of $f'$, the x-axis, the y-axis, and the line $x = 6$.

The area of region $A$ is 12, the area of region $B$ is 6.75 and the area of region $C$ is 6.75.

Let $g(x) = {\left( {f(x)} \right)^2}$. Given that $f'(6) = 16$, find the equation of the tangent to the graph of $g$ at the point where $x = 6$.

METHOD 1

$f'(5) = 0$     (A1)

valid reasoning including reference to the graph of $f'$     R1

eg$\;\;\;f'$ changes sign from negative to positive at $x = 5$, labelled sign chart for $f'$

so $f$ has a local minimum at $x = 5$     AG     N0

Note:     It must be clear that any description is referring to the graph of $f'$, simply giving the conditions for a minimum without relating them to $f'$ does not gain the R1.

METHOD 2

$f'(5) = 0$     A1

valid reasoning referring to second derivative     R1

eg$\;\;\;f''(5) > 0$

so $f$ has a local minimum at $x = 5$     AG     N0

[2 marks]

attempt to find relevant interval     (M1)

eg$\;\;\;f'$ is decreasing, gradient of $f'$ is negative, $f'' < 0$

$2 < x < 4\;\;\;$(accept “between 2 and 4”)     A1     N2

Notes:     If no other working shown, award M1A0 for incorrect inequalities such as $2 \le$ $x$ $\le$ 4, or “from 2 to 4”

[2 marks]

METHOD 1 (one integral)

correct application of Fundamental Theorem of Calculus     (A1)

eg$\;\;\;\int_0^6 {f'(x){\text{d}}x = } f(6) - f(0),{\text{ }}f(6) = 14 + \int_0^6 {f'(x){\text{d}}x}$

attempt to link definite integral with areas     (M1)

eg$\;\;\;\int_0^6 {f'(x){\text{d}}x =  - 12 - 6.75 + 6.75,{\text{ }}\int_0^6 {f'(x){\text{d}}x = {\text{Area }}A + {\text{Area }}B + {\text{ Area }}C} }$

correct value for $\int_0^6 {f'(x){\text{d}}x}$     (A1)

eg$\;\;\;\int_0^6 {f'(x){\text{d}}x}  =  - 12$

correct working     A1

eg$\;\;\;f(6) - 14 =  - 12,{\text{ }}f(6) =  - 12 + f(0)$

$f(6) = 2$     A1     N3

METHOD 2 (more than one integral)

correct application of Fundamental Theorem of Calculus     (A1)

eg$\;\;\;\int_0^2 {f'(x){\text{d}}x}  = f(2) - f(0),{\text{ }}f(2) = 14 + \int_0^2 {f'(x)}$

attempt to link definite integrals with areas     (M1)

eg$\;\;\;\int_0^2 {f'(x){\text{d}}x}  = 12,{\text{ }}\int_2^5 {f'(x){\text{d}}x =  - 6.75} ,{\text{ }}\int_0^6 {f'(x)}  = 0$

correct values for integrals     (A1)

eg$\;\;\;\int_0^2 {f'(x){\text{d}}x}  =  - 12,{\text{ }}\int_5^2 {f'(x)} {\text{d}}x = 6.75,{\text{ }}f(6) - f(2) = 0$

one correct intermediate value     A1

eg$\;\;\;f(2) = 2,{\text{ }}f(5) =  - 4.75$

$f(6) = 2$     A1     N3

[5 marks]

correct calculation of $g(6)$ (seen anywhere)     A1

eg$\;\;\;{2^2},{\text{ }}g(6) = 4$

choosing chain rule or product rule     (M1)

eg$\;\;\;g'\left( {f(x)} \right)f'(x),{\text{ }}\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{\text{d}}y}}{{{\text{d}}u}} \times \frac{{{\text{d}}u}}{{{\text{d}}x}},{\text{ }}f(x)f'(x) + f'(x)f(x)$

correct derivative     (A1)

eg$\;\;\;g'(x) = 2f(x)f'(x),{\text{ }}f(x)f'(x) + f'(x)f(x)$

correct calculation of $g'(6)$ (seen anywhere)     A1

eg$\;\;\;2(2)(16),{\text{ }}g'(6) = 64$

attempt to substitute their values of $g'(6)$ and $g(6)$ (in any order) into equation of a line     (M1)

eg$\;\;\;{2^2} = (2 \times 2 \times 16)6 + b,{\text{ }}y - 6 = 64(x - 4)$

correct equation in any form     A1     N2

eg$\;\;\;y - 4 = 64(x - 6),{\text{ }}y = 64x - 380$

[6 marks]

[Total 15 marks]