I want to prove that $f: (-\pi/2,\pi/2) \to \mathbb{R}: x \mapsto e^x \tan x$ is strictly monotonically increasing on the given interval $D = (-\pi/2,\pi/2)$. I'd do this with the condition that $$ f'(x) > 0 \, \forall x \in D \Rightarrow f \text{ strictly mono. incr.} $$ See Q&A answer below. It's a trivial question, really, but I already wrote it all out, so here it goes.
Prove that $e^x \tan x$ is strictly monotonically increasing on $(-\pi/2, \pi/2)$
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real-analysis
derivatives
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0Both function are individually monotonically increasing – 2017-01-05
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2@jnyan: "Individually monotonically increasing" is not enough: $f(x) = g(x) = x$ are increasing everywhere, but their product is not increasing on any interval containing a negative number. ;) (_Positive and increasing_ suffices, but the tangent function isn't positive in the stated interval, so work is required.) – 2017-01-05
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0I am sorry. I meant one is positive always and monotonically increasing. Won't it work then? – 2017-01-05
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0@jnyan $f(x)=x$ and $g(x)=x+1\ge 0$ on $[-1,0]$. – 2017-01-05
2 Answers
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So, after writing my question down, I already found the answer myself, so this is gonna be a Q&A answer.
Taking the derivative, we have $$ \frac{d}{dx}f(x) = e^x \left( \tan x + \frac{1}{\cos^2 x}\right) $$ Now, this product is obviously bigger than 0 if both factors are positive or both are negative. $e^x$ is always positive for $x \in \mathbb{R}$, so that means we must show that $(\tan x + \frac{1}{\cos^2 x}$) is positive.
This is where I was stuck. I then did the following steps: $$ \begin{align} \tan x + \frac{1}{\cos^2 x} &>0 \\ \tan x &> -\frac{1}{\cos^2 x} \\ \frac{\sin x}{\cos x} &> -\frac{1}{\cos^2 x} \\ \sin x \cdot \cos x &> -1 \end{align} $$ Now we know that for $x \in D, 0<\cos x \leq 1$ and $-1 < \sin x < 1$, and the product of these will always be $>-1$. Q.E.D.
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1That's about 90% correct. When you say "so this means that ... must be positive", you should instead write "so this means that we must show that ... is positive." Similarly, when you start with the trig inequality and simplify it to something true, you should put double-implication signs between every two adjacent steps. The third step, where you go from $\cos^2$ to $\cos$ also requires the justification that you've put one line further down, because you've multiplied through by $\cos$ there as well. Ask yourself, though: why doesn't this same proof work for $-\pi < x <\pi$? Or does it? – 2017-01-05
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1(Note that you can combine the last two steps into one by multiplying through by $\cos^2$, which is always positive, so that simplifies things a little.) – 2017-01-05
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1If you use $\tan'x=1+\tan^2x$ then $t^2+t+1>0$ becomes trivial. – 2017-01-05
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by Maclaurin Series we have: e^x=1+x/1!+x^2/2!+x^3/3!+⋯,-∞
from this if 0 < x < y then e^x < e^y and it follows that e^(-y) < e^(-x). hence e^x is strictly increasing on the whole real axis.