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After finding the first derivative which is $y'=3x^2 + 4x -1$, I found the two $x$ values after using the quadratic formula to factor the above when $y'=0$ which were $x = 0.22$ and $x = -1.55$. From here, I found that the maximum occurs at $(-1.55, 8.63)$ and that the minimum occurs at $(0.22, 5.89)$.

Now these values I found would they be the local extreme values? If so how would I find the absolute values?

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  • $\begingroup$ There are no absolute extreme values of that function. Absolute extreme values are the highest or lowest points that the function ever reaches. However, because the highest exponent of that function is odd, it never reaches any highest or lowest points. $\endgroup$ – Polygon Jul 23 '16 at 1:02
  • $\begingroup$ Thanks, I totally understand now! $\endgroup$ – Jessica Jul 23 '16 at 1:06
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Are any boundaries given for $x$? If not, consider $x\rightarrow \infty$ and $x \rightarrow -\infty$. We see that $y \rightarrow \infty$ or $-\infty$. These would be the absolute extreme values.

If boundaries are given, evaluate $y$ at the borders. Ex: if $x\in [0, 2]$, evaluate for $x=0$ and $x=2$ and compare these values with the local extrema you found earlier.

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