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When asked to find the interval on which the following curve is concave upward $$ y = \int_0^x \frac{1}{94+t+t^2} \ dt $$

What is basically being asked to be done here? Evaluate the integral between $[0,x]$ for some function and then differentiate twice to find the concavity of the resulting function?

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    $\begingroup$ Do you know that $y''\leq 0$ tells a lot about convexity? Are you able to prove that $y''\leq 0$ through the fundamental theorem of Calculus? Hint: you do not need to compute the integral in a explicit way. $\endgroup$ – Jack D'Aurizio Jun 21 '17 at 21:02
  • $\begingroup$ Hint: $y = y(x)$. How do we see if a function $y$ is concave up? $\endgroup$ – Sean Roberson Jun 21 '17 at 21:04
  • $\begingroup$ Set the second derivative equal to zero? $\endgroup$ – Computer Jun 21 '17 at 21:06
  • $\begingroup$ Apply the fundamental theorem of calculus. It is "fundamental," you should know it. $\endgroup$ – Doug M Jun 21 '17 at 21:09
  • $\begingroup$ a function is concave if $f(\frac{x+x'}2)\geq f(x)/2+f(x')/2$. Now use the lineartiy of the integral operator $\endgroup$ – tired Jun 21 '17 at 21:09
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Hints:

$$y'=\frac1{x^2+x+94}\;,\;\;y''=-\frac{2x+1}{(x^2+x+94)^2}$$

and if $\;x>0\;$ , then the sign of $\;y'' \;$ is...

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  • $\begingroup$ Actually you should have said if $x>-\frac12$ $\endgroup$ – Mark Fischler Jun 21 '17 at 21:11
  • $\begingroup$ @MarkFischler That's for the OP to deduce. I just observed that the given integral is $\;\int_0^x\;$ and thus assumed it is likely that $\;x>0\;$ ... $\endgroup$ – DonAntonio Jun 21 '17 at 21:12

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