Functions (Transformations and Inverses)
Showing 1-3 of 3 problems
- Sketch the curve \(y = xe^x\), giving the coordinates of any stationary points.
- The function \(f\) is defined by \(f(x) = xe^x\) for \(x \geqslant a\), where \(a\) is the minimum possible value such that \(f\) has an inverse function. What is the value of~\(a\)? Let \(g\) be the inverse of \(f\). Sketch the curve \(y = g(x)\).
- For each of the following equations, find a real root in terms of a value of the function~\(g\), or demonstrate that the equation has no real root. If the equation has two real roots, determine whether the root you have found is greater than or less than the other root.
- \(e^{-x} = 5x\)
- \(2x \ln x + 1 = 0\)
- \(3x \ln x + 1 = 0\)
- \(x = 3\ln x\)
- Given that the equation \(x^x = 10\) has a unique positive root, find this root in terms of a value of the function~\(g\).
The point \(P\) has coordinates \((x,y)\) which satisfy \[ x^2+y^2 + kxy +3x +y =0\,. \]
- Sketch the locus of \(P\) in the case \(k=0\), giving the points of intersection with the coordinate axes.
- By factorising \(3x^2 +3y^2 +10xy\), or otherwise, sketch the locus of \(P\) in the case \(k=\frac{10}{3}\,\), giving the points of intersection with the coordinate axes.
- In the case \(k=2\), let \(Q\) be the point obtained by rotating \(P\) clockwise about the origin by an angle~\(\theta\), so that the coordinates \((X,Y)\) of \(Q\) are given by \[ X=x\cos\theta +y\sin\theta\,, \ \ \ \ Y= -x\sin\theta + y\cos\theta\,. \] Show that, for \(\theta =45^\circ\), the locus of \(Q\) is \( \sqrt2 Y= (\sqrt2 X+1 )^2 - 1 .\) Hence, or otherwise, sketch the locus of \(P\) in the case \(k=2\), giving the equation of the line of symmetry.
- \(k = 0\), we have \(x^2 + y^2 + 3x + y = 0\), ie \((x+\tfrac32)^2+(y+\tfrac12)^2 = \frac{10}{4}\).
- \(3x^2 + 3y^2 +10xy = (3x+y)(x+3y)\) so \(x^2 + y^2 + \tfrac{10}3xy + 3x+y = (3x+y)(\frac{x+3y}{3}+1) = 0\) so we have the line pair \(3x +y =0\), \(x+3y + 3 = 0\)
- If \(k = 2\) then \((x+y)^2 + (x+y)+2x = 0\). If \(\theta = 45^\circ\) then \( X = \frac1{\sqrt{2}}(x+y), Y = \frac{1}{\sqrt{2}}(y-x)\), ie \(x+y = \sqrt{2}X\) and \(x = \frac{1}{\sqrt2}(X-Y)\), so our equation is:
\begin{align*}
0 &= 2X^2 + \sqrt{2}X + \sqrt{2}(X-Y) \\
&= (\sqrt{2}X + 1)^2 - 1 - \sqrt{2} Y
\end{align*}
which would be a parabola with line of symmetry \(X = -\frac{1}{\sqrt{2}}\).
However, we are actually looking at that parabola rotated by \(45^\circ\) anticlockwise.
- The functions \(\mathrm{a, b, c}\) and \(\mathrm{d}\) are defined by
- \({\rm a}(x) =x^2 \ \ \ \ (-\infty < x < \infty),\)
- \({\rm b}(x) = \ln x \ \ \ \ (x > 0),\)
- \({\rm c}(x) =2x \ \ \ \ (-\infty < x < \infty),\)
- \({\rm d}(x)= \sqrt x \ \ \ \ (x\ge0) \,.\)
- The functions \(\mathrm{f}\) and \(\mathrm{g}\) are defined by
- \(\f(x)= \sqrt{x^2-1\,} \ \ \ \ (\vert x \vert \ge 1),\)
- \(\g(x) = \sqrt{x^2+1\,} \ \ \ \ (-\infty < x < \infty).\)
- Sketch the graphs of the functions \(\h\) and \({\rm k}\) defined by
- \(\h(x) = x+\sqrt{x^2-1\,}\, \ \ \ \ ( x \ge1)\),
- \({\rm k}(x) = x-\sqrt{x^2-1\,}\, \ \ \ \ (\vert x\vert \ge1),\)
- \begin{align*} cb(x) &= c(b(x)) \\ &= 2 \ln x \quad (x > 0) \\ ab(x) &= (b(x))^2 \\ &= (\ln x)^2 \quad (x > 0) \\ da(x) &= \sqrt{a(x)} \\ &= \sqrt{x^2} \\ &= |x| \quad (-\infty < x < \infty) \\ ad(x) &= (d(x))^2 \\ &= (\sqrt{x})^2 \\ &= x \quad (x \geq 0) \end{align*} The domains are specified above. The ranges are \(\mathbb{R}, \mathbb{R}_{\geq 0}, \mathbb{R}_{\geq 0}, \mathbb{R}_{\geq 0}\) respectively.
- \begin{align*} fg(x) &= \sqrt{g(x)^2-1} \quad (|g(x)| \geq 1) \\ &= \sqrt{x^2+1-1} \\ &= |x| \end{align*} So \(fg: \mathbb{R} \to \mathbb{R}_{\geq 0}\). \begin{align*} gf(x) &= \sqrt{f(x)^2 + 1} \\ &= \sqrt{\left ( \sqrt{x^2-1} \right)^2+1} \quad (|x| \geq 1) \\ &= |x| \end{align*} So \(gf: \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}\)
