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Let $$f(x)=x^4+4x^3+100$$ Find the global minimum value of the function $f$, that is, give the minimum value of $f$ and the value of $x$ for which this occurs.

This is not a homework problem. I have a exam and I don't know how to answer this question. I did the first and second derivative test. I don't need answer. I don't wanna fail my exam. Can anyone explain step by step how they would do these kind of problems. I am sorry if this is a lot to explain, but I looked on youtube and all of the videos explained with the intervals. It did not help. Thank you.

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  • $\begingroup$ When you did the first derivative test, what did you get to be the critical points? $\endgroup$ – Mike Pierce Nov 28 '14 at 16:49
  • $\begingroup$ i got x=0 and x=-3 @mapierce271 $\endgroup$ – Kmi Kshetri Nov 28 '14 at 16:50
  • $\begingroup$ try to post here openstudy.com ,all the best $\endgroup$ – Jonas Kgomo Nov 28 '14 at 16:50
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    $\begingroup$ $f(x)$ tends to $\infty$ as $x$ tends to either $\infty$ or $-\infty$. The smallest local minimum will therefore be the global minimum. $\endgroup$ – David Mitra Nov 28 '14 at 16:50
  • $\begingroup$ @DavidMitra but there two minimum and I don't know my intervals. could explain this step by step? $\endgroup$ – Kmi Kshetri Nov 28 '14 at 16:54
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You need to find the local minimums of the function. They can be only in the points where $f'(x) = 0$. $f'(x) = 4x^3 + 12x^2 = 4x^2(x + 3)$. So the two zeroes are $x_1 = 0$ and $x_2 = -3$. There is one more requirement for a point $x$ to be local minimum. It is for all $y$ a little less than $x$ the function to be decreasing and for all $y$ a little bigger than $x$ the function to be increasing. So we want $f'(y) < 0$ for $y$ in some really small interval $(-\epsilon, x)$, where $\epsilon > 0$ and $f'(y) > 0$ for $y \in (x, \epsilon)$. $x_1 = 0$ is obviously not such point, but $x_2 = -3$ is OK. So the minimum of $f$ has to be $f(-3) = 73$.

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show that $$x^4+4x^3+100\geq 73$$ for all real $x$ since $$(x^2-2x+3)(x+3)^2\geq 0$$

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