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$f(x) = e^x - x^2/2 - x + \ln x$

The answer given by my book is that $f(x)$ is strictly increasing for $x\in(0,+οο)$.

But it's giving me trouble. I tried:

$f'(x) = e^x - x- 1 + 1/x$ but I have no idea how to prove $f'(x)>0$.

Any help is appreciated.

Julián Aguirre
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alexk745
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  • can you find the zeros of $f'.$ you may need to use the solver function on your calculator. i t has only one zero at $-1.454$ there is also the vertical asymptote at $x = 0$ the rest is sign analysis of $f'$ – abel Mar 25 '15 at 18:27

3 Answers3

1

Here are two possibilities.

  1. From Taylor series, we know that $e^x = 1 + x + x^2/2 + \text{(other positive terms)}$. Subtracting $x + 1$ leaves $\text{positive terms} + \frac{1}{x}$, which is positive on $(0, \infty)$.

  2. We might also proceed naively from first principles. Note that $f'(1) = e + 1 > 0$, and $f''(x) = e^x - 1 - 1/x^2$. For $x > 1$, we have that $e^x > 2 > 1 + \frac{1}{x^2}$. Thus since $f''(x) > 0$ for $x > 1$ and $f'(1) > 0$, the derivative $f'(x)$ is increasing and positive on $(1,\infty)$. All that's left is to consider the interval $(0,1)$.

    In this interval, since $\frac{1}{x} > x$ and $e^x > 1$, we get the desired inequality. Patching them together gives your answer. $\diamondsuit$

Community
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davidlowryduda
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See from $x \in [0,1/2]$ $1/x>1+x$ and for $x \in (1/2, \infty)$ $e^x+1/x> 1+x$ so we are done.

Ri-Li
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$$f'(x)=e^x-x-1+\frac1{x}$$ $$\frac 1{x} >0$$ Also,you can find equation of tangent to $e^x$ at $x=0$

It's equation is $y=1+x$ you can easily find that this line is below the curve for all values of x.Thus,$e^x>1+x$

Thus,$f'(x)>0$

Akshay Bodhare
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