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Find the critical points, domain endpoints, and local extreme values for the function: $$\displaystyle y=\begin{cases} -x^2-4x+5 & \text{if }\;\;\; x \leq 1\\ -x^2+5x-4 & \text{if }\;\;\; x>1\\ \end{cases}$$

I found the local maximum: $(-2,9),\left(\dfrac{5}{2},\dfrac{9}{4}\right)$

I don't know how to find the local minimum for this problem. The answer for the local min is $(1,0),$ and I don't see how to get that.

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From your previous calculations, you can see that there are only two points for which $f'(y) = 0$, and both correspond to local maxima via the second derivative test. Furthermore, although the function is defined for all reals, there is no global minimum on $x \in (-\infty, \infty)$, because, for instance, $y \to -\infty$ as $x \to +\infty$. So, the only other possibility for a local extremum is at a non-differentiable point; i.e., when $x = 1$. This corresponds to a cusp: clearly, $y = 0$ at $x = 1$, and $y$ is continuous at this point. So all that remains is to show that for a sufficiently small neighborhood around $x = 1$, $f(y) \ge 0$. This is easy to see by returning to the derivative around $1$ and seeing from its sign that $f$ is decreasing for $x < 1$ and increasing for $x > 1$.

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  • $\begingroup$ thanks, but what do you mean by $f(y) \geq 0$? $\endgroup$ – Emi Matro Jan 5 '14 at 0:50
  • $\begingroup$ do you mean $f(x) \geq f(c),$ $c$ being the critical point $x=c$, in this case $1$? $\endgroup$ – Emi Matro Jan 5 '14 at 0:52
  • $\begingroup$ I meant $f(x) \ge 0$ for all $x$ in a sufficiently small neighborhood of $1$. $\endgroup$ – heropup Jan 5 '14 at 0:58
  • $\begingroup$ ok thanks, but why $f(x) \geq 0$? why $\geq 0$? $\endgroup$ – Emi Matro Jan 5 '14 at 1:01
  • $\begingroup$ Because $f(1) = 0$, and you want to show that no other point in the neighborhood is smaller than this. Actually, you want to establish that if $x \ne 1$, then the inequality is strict, again for a sufficiently small neighborhood. $\endgroup$ – heropup Jan 5 '14 at 1:05

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