Standard +0.3 This question requires applying the product rule twice to find the second derivative, then solving a simple equation. While it involves multiple steps (first derivative, second derivative, solving), each step is straightforward application of standard techniques with no conceptual difficulty or novel insight required. The algebra is manageable and the exponential function makes the equation easy to solve. Slightly above average due to the two-derivative requirement, but still a routine calculus exercise.
6 Find the exact coordinates of the point on the curve \(y = x \mathrm { e } ^ { - \frac { 1 } { 2 } x }\) at which \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 0\).
At any stage, state the correct derivative of \(e^{-\frac{1}{x}}\) or \(e^{\frac{1}{x}}\)
B1
Use product or quotient rule
M1
Obtain correct first derivative in any form
A1
Obtain correct second derivative in any form
B1 \(\checkmark\)
Equate second derivative to zero and solve for \(x\)
M1
Obtain \(x = 4\)
A1
Obtain \(y = 4e^{-\frac{1}{4}}\), or equivalent
A1
[7]
At any stage, state the correct derivative of $e^{-\frac{1}{x}}$ or $e^{\frac{1}{x}}$ | B1 |
Use product or quotient rule | M1 |
Obtain correct first derivative in any form | A1 |
Obtain correct second derivative in any form | B1 $\checkmark$ |
Equate second derivative to zero and solve for $x$ | M1 |
Obtain $x = 4$ | A1 |
Obtain $y = 4e^{-\frac{1}{4}}$, or equivalent | A1 | [7]
6 Find the exact coordinates of the point on the curve $y = x \mathrm { e } ^ { - \frac { 1 } { 2 } x }$ at which $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 0$.
\hfill \mbox{\textit{CAIE P2 2008 Q6 [7]}}