Consider the functions f and g defined by $f(x) = {2^{\frac{1}{x}}}$ and $g(x) = 4 - {2^{\frac{1}{x}}}$ , $x \ne 0$ .

(a)     Find the coordinates of P, the point of intersection of the graphs of f and g .

(b)     Find the equation of the tangent to the graph of f at the point P.

(a)     ${2^{\frac{1}{x}}} = 4 - {2^{\frac{1}{x}}}$

attempt to solve the equation     M1

x = 1     A1

so P is (1, 2) , as $f(1) = 2$     A1     N1

(b)     $f'(x) = - \frac{1}{{{x^2}}}{2^{\frac{1}{x}}}\ln 2$     A1

attempt to substitute x-value found in part (a) into their $f'(x)$     M1

$f'(1) = - 2\ln 2$

$y - 2 = - 2\ln 2(x - 1)\,\,\,\,\,{\text{(or equivalent)}}$     M1A1     N0

[7 marks]

Most candidates answered part (a) correctly although some candidates showed difficulty solving the equation using valid methods. Part (b) was less successful with many candidates failing to apply chain rule to obtain the derivative of the exponential function.