f(x) = lnx/x

(a): Find f'(x) and show that x = e is a maximum
(b): Deduce that e^x >= x^e, for x > 0

1 Answer

  • [tex]f(x) = \frac{lnx}{x}[/tex]

    (a) [tex]f'(x) = \frac{d}{dx}[\frac{lnx}{x}][/tex]
    Using the quotient rule:
    [tex]f'(x) = \frac{\frac{1}{x} \cdot x - lnx}{x^{2}}[/tex]
    [tex]f'(x) = \frac{1 - lnx}{x^{2}}[/tex]

    For maximum, f'(x) = 0;
    [tex]\frac{1 - lnx}{x} = 0[/tex]
    [tex]lnx = 1, x = e[/tex]

    (b) Deduce:
    [tex]e^{x} \geq x^{e}, x > 0[/tex]

    Since x = e is the greatest value, then f(e) ≥ f(x) > f(0)
    [tex]\frac{ln(e)}{e} \geq \frac{lnx}{x}[/tex]
    [tex]\frac{1}{e} \geq \frac{lnx}{x}[/tex], since ln(e) is simply equal to 1

    Now, since x > 0, then we don't have to worry about flipping the signs when multiplying by x.
    [tex]\frac{x}{e} \geq lnx[/tex]
    [tex]x \geq elnx[/tex]
    [tex]x \geq ln(x^{e})[/tex]

    Taking the exponential to both sides will cancel with the natural logarithmic function in the right hand side to produce:
    [tex]e^{x} \geq e^{ln(x^{e}})[/tex]
    [tex]e^{x} \geq x^{e}[/tex], as required.