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Find the extreme values of the function $f(x)=2^x-x$ in its domain.

$f'(x)=2^xln(2)-1$ $f'(x)=0 \implies x=-{\frac {\ln \left( \ln \left( 2 \right) \right) }{\ln \left( 2 \right) }}$ then we have ${2}^{-{\frac {\ln \left( \ln \left( 2 \right) \right) }{\ln \left( 2 \right) }}}+{\frac {\ln \left( \ln \left( 2 \right) \right) }{\ln \left( 2 \right) }}={\frac {1+\ln \left( \ln \left( 2 \right) \right) }{\ln \left( 2 \right) }} $

But answer in the book is $\frac{e\log(2)}{2}$

  • 0
    Maybe the solution of the book is wrong. – 2017-01-30
  • 0
    There is no need to calculate to see, that the book contains a wrong value for the minimum: A simple plot shows you that the minimum is between $0.91$ and $0.92$ . – 2017-01-30

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Textbook authors are incapable of making errors in anything they write. This is because they have made a bargain with devil before beginning their book.

However, the solutions to textbook exercises are not written by textbook authors. They are written by graduate students who have accepted the task because it sounded better than the alternative of tutoring undergraduates in calculus.

The answers in a textbook can be wrong. In this case, you are right and the book is wrong.