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I am struggling with problems like this, just trying to grasp the concept. Using the Intermediate Value Theorem and Rolle's Theorem, prove that $4x^5+x^3+2x+1=0$ has exactly one real root.

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$f'(x) = 20x^4 + 3x^2 + 2 > 0$ for all real values $x$, and $f(0) = 1 > 0$, $f(-1) = -6 < 0$, then by the intermediate value theorem there exits one and only one real root in $(-1,0)$ for the equation: $4x^5 + x^3 + 2x + 1 = 0$

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Function $$f(x)=4x^5+x^3+2x+1$$ varies from $-\infty$ to $+\infty$ beacause of the higher power and corresponding sign for its coefficient.

Its derivative $$f'(x)=20 x^4+3 x^2+2$$ does not show any real root ($f'(x)=0 $ results in a quadratic equation in $y=x^2$) and it is the always positive. So, $\cdots$

I am sure that you can take from here.

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  • $\begingroup$ At the risk of sounding incompetent, I actually can't. If possible, and I know it would be a pain for you to do so, I'd like to see the logic in this broken down step by step. I have a very terrible calculus professor and I've been having to (almost) completely self-educate. Regardless, thanks for the help. $\endgroup$ – DrakkorNoir Oct 30 '14 at 4:56
  • $\begingroup$ By the way, welcome to Math SE ! Just think about a function whic permanently increases (since its derivative is always positive). How many times do you think the graph will intersect the $x$ axis ? Is this any better ? Just post if not. $\endgroup$ – Claude Leibovici Oct 30 '14 at 4:59
  • $\begingroup$ Okay, that makes sense. It'll never be 0 again since it keeps increasing. My question is, how did we prove that it even had a root? $\endgroup$ – DrakkorNoir Oct 30 '14 at 5:01
  • $\begingroup$ No; the function starts from $-\infty$, so there will be one single intersection. You will have one root and , as OC-Sansoo showed in his/her answer, this root will be between $-1$ and $0$. $\endgroup$ – Claude Leibovici Oct 30 '14 at 5:04
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Alternatively, suppose $x(4(x^2)^2+(x^2)+2) = -1$. Since the term in parentheses is strictly positive, we see that any real root must be strictly negative and, of course, satisfies $4(x^2)^2+(x^2)+2 = -{1 \over x}$.

The function $f(y) = 4(y^2)^2+(y^2)+2 + {1 \over y}$ is strictly decreasing on $(-\infty, 0)$, $\lim_{y\to -\infty} f(y) = \infty$ and $\lim_{y \uparrow 0} f(y) = -\infty$, hence by the intermediate value theorem there is exactly one $y \in (0,\infty)$ such that $f(y) = 0$.

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Let $f(x) = 4x^{5} + x^{3} + 2x + 1$. As noted in other answer there is a real root of $f(x)$ in interval $(-1, 0)$ because $f(0)$ and $f(-1)$ are of opposite signs. But I would like to present a rather elementary argument for the uniqueness of the root. Clearly we can see that if $x > 0$ then $f(x) > 0$, so that the root if any must be negative. Let's assume on the contrary that there are two distinct roots of $f(x)$ namely $a, b$ and $a \neq b$ and both $a, b$ are negative. Then $f(a) = 0 = f(b)$. So we have $f(a) - f(b) = 0$. This means that $$\begin{aligned}4(a^{5} - b^{5}) + (a^{3} - b^{3}) + 2(a - b) &= 0\\ \Rightarrow 4(a^{4} + a^{3}b + a^{2}b^{2} + ab^{3} + b^{4}) + (a^{2} + ab + b^{2}) + 2 &= 0\end{aligned}$$ Now we can see that both $a, b$ are negative and hence each term of LHS is positive and hence can't sum to $0$ which is on RHS. This contradiction shows that we can't have two distinct roots $a, b$ of $f(x)$.

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The derivative tells you that the function is strictly increasing. You need to find $a$ and $b$ such that $f(a)<0$ and $f(b)>0$. Taking into account the intermediate value theorem and that the function is strictly increasing you are able to conclude that the function has a real root on the interval $(a,b)$ and that it is unique. Otherwise the function would not be strictly increasing.

By the way, the root is $x\approx-0.430521$.

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By Descartes’ Law of Signs, there are no positive roots and 1 negative root.

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