2014 ONE(d)

Merit
Question
Find the xx-value at which the tangent to the function y=4e2x−2+8xy = \dfrac{4}{e^{2x-2}} + 8x is parallel to the xx-axis.

*Show any derivatives that you need to find when solving this problem.*
Official Answer
y=4e2x−2+8x=4e−2x+2+8xy=\dfrac{4}{e^{2x-2}}+8x=4e^{-2x+2}+8x

dydx=−8e−2x+2+8\dfrac{dy}{dx}=-8e^{-2x+2}+8

Parallel to xx-axis ⇒dydx=0\Rightarrow \dfrac{dy}{dx}=0

8e−2x+2=88e^{-2x+2}=8

e−2x+2=1e^{-2x+2}=1

−2x+2=0-2x+2=0

x=1x=1
Grading Criteria

Achievement (u)

  • A correct expression for dydx\dfrac{dy}{dx}.

Merit (r)

  • A correct solution.

Excellence T1

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Excellence T2

-
Exam Paper (2014)Assessment Schedule (2014)
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