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Find the intervals where the function is convex and concave. $$f (x) = e^{2x} - 2e^x$$

I tried differentiating twice, and my answer is: concave when $x < \ln (1/2)$ and convex when $x > \ln (1/2)$. However the key says the other way around...

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    $\begingroup$ did you tried differentiaing twice $\endgroup$ – RE60K Feb 19 '15 at 15:49
  • $\begingroup$ Yes and my answer is: concave when x < ln (1/2) and convex when x > ln (1/2). However the key says the other way around... $\endgroup$ – Casper Lindberg Feb 19 '15 at 15:52
  • $\begingroup$ @CasperLindberg Be aware some books assign the names concave and convex in many different ways. What is important is to know what shape the graph has. $\endgroup$ – Tom Feb 19 '15 at 15:54
  • $\begingroup$ I'm following the book's definition. $\endgroup$ – Casper Lindberg Feb 19 '15 at 16:00
  • $\begingroup$ You correctly found the point where the second derivative of the function changes its sign. Now, what you call each half is irrelevant, really. However, if rain falls from above, then $x>\ln1/2$ will collect water because it is concave. $\endgroup$ – wjm Feb 19 '15 at 16:05
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$f'(x) = 2e^{2x} - 2e^x \to f''(x) = 4e^{2x} - 2e^x$. $f$ is concave if $f''(x) < 0$, and is convex if $f''(x) > 0$. $f''(x) < 0 \iff 2e^x\left(2e^x - 1\right)<0 \iff 2e^x - 1 < 0 \iff e^x < \dfrac{1}{2} \iff x < \ln\left(\dfrac{1}{2}\right)=-\ln 2$

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