$$f(x) = 4x^5 + x^3 + 7x - 2$$
$f'(x)>0$ for all $x$, But $f''(x)=80x^3+6x$ can be negative for negative $x$, so is the function strictly increasing?
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This function has one real root on the interval $[0,1].$ – Chickenmancer May 04 '17 at 00:58
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Since $f^\prime$ has no zero in $\Bbb R$, By Rolle there has to be only one zero at max. Since $f$ has odd degree, therefore only one zero. – May 04 '17 at 01:31
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Since $f'(x)=x^2(20x^2+3)+7$ is positive, thus function is definitely increasing. Getting a negative $f"(x)$ is nothing but getting a local maxima in increasing function
and $f(0)=-2$ while $f(1)>0$. Thus this shows that this function has a real root in $(0,1)$ and this is only real root for $x>0$
The Dead Legend
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So the second derivative does not make a difference in it being strictly increasing? – onelessproblem May 04 '17 at 00:59
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1@onelessproblem The concavity of a function does not dictate wheather it's increasing or not. For example, $e^x$ is concave up and increasing. $\ln x$ is concave down and increasing. – Ovi May 04 '17 at 01:02
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Also, I'd suggest you to notice that you can get a local maxima/minima only at critical points. which, over here is $x=0$ @onelessproblem – The Dead Legend May 04 '17 at 01:02
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