Question 1(d) Merit

A chain-rule root, a product-rule tangent, a decreasing-interval question mixing a square with a log, stationary points of an exponential product, and a tangent at a point of inflection.

Find the x-value(s) of any stationary points on the graph of y = (2x − 1)e^−2x, and determine their nature.

The 2024 paper, page 3 ↗ Marking schedule, page 1 ↗

Work it out

Try it on paper first. Stuck? Take one hint at a time.

1. 1 Product rule: what are u and v, and their derivatives?

   u = 2x − 1, u' = 2
   v = e^−2x, v' = −2e^−2x

2. 2 Differentiate, then take out the common factor.

   dy/dx = 2e^−2x − 2(2x − 1)e^−2x = e^−2x(4 − 4x)

3. 3 Set it to zero.

   e^−2x is never zero, so 4 − 4x = 0 and x = 1.

4. 4 Is it a maximum or a minimum? Use the second derivative.

   (d²y)/dx² = −2e^−2x(4 − 4x) + e^−2x(−4) = e^−2x(8x − 12)
   At x = 1: −4e⁻² ≈ −0.541 < 0, so a maximum.

What the marker wanted

• Achieved A correct dy/dx.
• Merit x = 1, shown to be a maximum by a calculus method, from a correct derivative.

In our words, from NZQA’s 2024 marking schedule.

How did it go?

Got it moves it further out. Review brings it back tomorrow.