Exponential decay curve meeting y equals x, a tangent of gradient minus one and areas bounded by the curve, its inverse and the tangent

Exponential decay curve meeting y equals x, a tangent of gradient minus one and areas bounded by the curve, its inverse and the tangent

15 marks

Let \(f(x)=60\mathrm{e}^{-\frac{x}{3}}\), \(x\in\mathbb{R}^+\). The line \(y=x\) intersects the curve \(y=f(x)\) at \(P\).

  1. Determine the \(x\)-coordinate of \(P\).[2]

The straight line \(L\), with gradient \(-1\), is a tangent to the curve \(y=f(x)\) at \(Q\).

  1. Find the coordinates of \(Q\), giving each in exact form.[4]
  2. Show that \(L\) has equation \(y=3+3\ln20-x\).[2]

The region \(R\), shaded in the diagram, is bounded by the graph of \(f\), the line \(y=x\) and the line \(L\).

    1. Determine where \(L\) and \(y=x\) intersect, giving the \(x\)-coordinate.
    2. Hence, calculate the area of \(R\).[5]

In addition, \(L\) is a tangent to the curve \(y=f^{-1}(x)\), where \(f^{-1}\) is the inverse function of \(f\), as shown in the second diagram.

  1. Calculate the area of the region shaded in the second diagram, enclosed by \(L\) and the curves \(y=f(x)\) and \(y=f^{-1}(x)\).[2]

Solution:

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Answer:(a) \(x=6.62\) (b) \(Q\left(3\ln20,\ 3\right)\) (c) \(y-3=-\left(x-3\ln20\right)\Rightarrow y=3+3\ln20-x\) (d)(i) \(x=\dfrac{3+3\ln20}{2}=5.99\) (d)(ii) \(1.30\text{ units}^2\) (e) \(2.60\text{ units}^2\)

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