2018 TWO(d)

Merit
Question
If y=ex(2x2−x−1)y = e^x(2x^2 - x - 1), find the value(s) of xx for which dydx=0\dfrac{dy}{dx} = 0.

You must use calculus and show any derivatives that you need to find when solving this
problem.
Official Answer
dydx=ex(2x2−x−1)+ex(4x−1)\dfrac{dy}{dx}=e^x(2x^2-x-1)+e^x(4x-1)
=ex(2x2+3x−2)=e^x(2x^2+3x-2)

dydx=0⇒ex(2x2+3x−2)=0\dfrac{dy}{dx}=0\Rightarrow e^x(2x^2+3x-2)=0

⇒2x2+3x−2=0\Rightarrow 2x^2+3x-2=0

⇒(x+2)(2x−1)=0\Rightarrow (x+2)(2x-1)=0

x=−2 or x=12x=-2\text{ or }x=\dfrac{1}{2}

Note ex=0e^x=0 has no solutions since ex>0 ∀x∈Re^x>0\ \forall x\in \mathbb{R}
Grading Criteria

Achievement (u)

  • Correct derivative.

Merit (r)

  • Correct solution with correct derivative.
  • Reference to ex=0e^x=0 not required.

Excellence T1

-

Excellence T2

-
Video Explanation
NCEA Level 3 Calculus Differentiation 2018 NZQA Exam - Worked Answers by infinityplusone(starts at 27:38)
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Exam Paper (2018)Assessment Schedule (2018)
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