Consider the curve with equation $f(x) = {{\text{e}}^{ - 2{x^2}}}{\text{ for }}x < 0$ .

Find the coordinates of the point of inflexion and justify that it is a point of inflexion.

METHOD 1

EITHER

Using the graph of $y = f'(x)$     (M1)

     A1

The maximum of $f'(x)$ occurs at x = −0.5 .     A1

OR

Using the graph of $y = f''(x)$.     (M1)

     A1

The zero of $f''(x)$ occurs at x = −0.5 .     A1

THEN

Note: Do not award this A1 for stating x = ±0.5 as the final answer for x .

$f( - 0.5) = 0.607( = {{\text{e}}^{ - 0.5}})$     A2

Note: Do not award this A1 for also stating (0.5, 0.607) as a coordinate.

EITHER

Correctly labelled graph of $f'(x)$ for $x < 0$ denoting the maximum $f'(x)$     R1

(e.g. $f'( - 0.6) = 1.17$ and $f'( - 0.4) = 1.16$ stated)     A1     N2

OR

Correctly labelled graph of $f''(x)$ for $x < 0$ denoting the maximum $f'(x)$     R1

(e.g. $f''( - 0.6) = 0.857$ and $f''( - 0.4) = - 1.05$ stated)     A1     N2

OR

$f'(0.5) \approx 1.21$. $f'(x) < 1.21$ just to the left of $x = - \frac{1}{2}$

and $f'(x) < 1.21$ just to the right of $x = - \frac{1}{2}$     R1

(e.g. $f'( - 0.6) = 1.17$ and $f'( - 0.4) = 1.16$ stated)     A1     N2

OR

$f''(x) > 0$ just to the left of $x = - \frac{1}{2}$ and $f''(x) < 0$ just to the right of $x = - \frac{1}{2}$     R1

(e.g. $f''( - 0.6) = 0.857$ and $f''( - 0.4) = - 1.05$ stated)     A1     N2

[7 marks] 

METHOD 2

$f'(x) = - 4x{{\text{e}}^{ - 2{x^2}}}$     A1

$f''(x) = - 4{{\text{e}}^{ - 2{x^2}}} + 16{x^2}{{\text{e}}^{ - 2{x^2}}}\,\,\,\,\,\left( { = (16{x^2} - 4){{\text{e}}^{ - 2{x^2}}}} \right)$     A1

Attempting to solve $f''(x) = 0$     (M1)

$x = - \frac{1}{2}$     A1

Note: Do not award this A1 for stating $x = \pm \frac{1}{2}$ as the final answer for x .

$f\left( { - \frac{1}{2}} \right) = \frac{1}{{\sqrt {\text{e}} }}{\text{ }}( = 0.607)$     A1

Note: Do not award this A1 for also stating $\left( {\frac{1}{2},\frac{1}{{\sqrt {\text{e}} }}} \right)$ as a coordinate.

EITHER

Correctly labelled graph of $f'(x)$ for $x < 0$ denoting the maximum $f'(x)$     R1

(e.g. $f'( - 0.6) = 1.17$ and $f'( - 0.4) = 1.16$ stated)     A1     N2

OR

Correctly labelled graph of $f''(x)$ for $x < 0$ denoting the maximum $f'(x)$     R1

(e.g. $f''( - 0.6) = 0.857$ and $f''( - 0.4) = - 1.05$ stated)     A1     N2

OR

$f'(0.5) \approx 1.21$. $f'(x) < 1.21$ just to the left of $x = - \frac{1}{2}$

and $f'(x) < 1.21$ just to the right of $x = - \frac{1}{2}$     R1

(e.g. $f'( - 0.6) = 1.17$ and $f'( - 0.4) = 1.16$ stated)     A1     N2

OR

$f''(x) > 0$ just to the left of $x = - \frac{1}{2}$ and $f''(x) < 0$ just to the right of $x = - \frac{1}{2}$     R1

(e.g. $f''( - 0.6) = 0.857$ and $f''( - 0.4) = - 1.05$ stated)     A1     N2

[7 marks]

Most candidates adopted an algebraic approach rather than a graphical approach. Most candidates found $f'(x)$ correctly, however when attempting to find $f''(x)$, a surprisingly large number either made algebraic errors using the product rule or seemingly used an incorrect form of the product rule. A large number ignored the domain restriction and either expressed $x = \pm \frac{1}{2}$ as the x-coordinates of the point of inflection or identified $x = \frac{1}{2}$ rather than $x = - \frac{1}{2}$. Most candidates were unsuccessful in their attempts to justify the existence of the point of inflection.