2020 TWO(c)

Merit
Question
Find the xx-coordinates of any stationary points on the graph of the function

f(x)=(2x3)ex2+kf(x) = (2x - 3)e^{x^2 + k}

You must use calculus and show any derivatives that you need to find when solving this
problem.
Official Answer
f(x)=(2x3)2xex2+k+2ex2+kf'(x)=(2x-3)2xe^{x^2+k}+2e^{x^2+k}
=ex2+k((2x3)2x+2)=e^{x^2+k}((2x-3)2x+2)
=ex2+k(4x26x+2)=e^{x^2+k}(4x^2-6x+2)
=2ex2+k(2x23x+1)=2e^{x^2+k}(2x^2-3x+1)

f(x)=02ex2+k=0f'(x)=0\Rightarrow 2e^{x^2+k}=0 or 2x23x+1=02x^2-3x+1=0

2ex2+k2e^{x^2+k} has no solutions since 2ex2+k2e^{x^2+k} is always positive.

2x23x+1=02x^2-3x+1=0

(2x1)(x1)=0(2x-1)(x-1)=0

x=12x=\frac{1}{2} or x=1x=1
Grading Criteria

Achievement (u)

  • Correct derivative.

Merit (r)

  • Correct solution with correct derivative.
  • Reference to 2ex2+k=02e^{x^2+k}=0 is not required

Excellence T1

-

Excellence T2

-
Video Explanation
NCEA Level 3 Calculus Differentiation 2020 NZQA Exam - Worked Answers by infinityplusone(starts at 23:51)
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