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.