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Show that the equation $x^3 +1 = 15x$ has three solutions in the interval $[-4,4]$

There are many elementary ways to solve this. But this question came out in exam for calculus!!

How do i solve it using calculus?

Here's what I learnt so far... Continuity, derivatives, mean value theorem, rolles theorem, fermats theorem, limits.

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3 Answers 3

up vote 2 down vote accepted

Let $f(x)=x^3-15x+1$; then $f\,'(x)=3x^2-15=3(x^2-5)=3(x-\sqrt5)(x+\sqrt5$. It’s easy to check that $f$ has a local maximum at $x=-\sqrt5$ and a local minimum at $x=\sqrt5$, either by the second derivative test, by examining the sign of the first derivative, or simply by knowing the shape of the graph of a cubic polynomial.

Now

$$\begin{align*} f(-4)&=-64+60+1=-3<0,\\ f(-\sqrt5)&=-5\sqrt5+15\sqrt5+1=10\sqrt5+1>0,\\ f(\sqrt5)&=5\sqrt5-15\sqrt5+1=-10\sqrt5+1<0,\text{ and}\\ f(4)&=64-60+1=5>0\;, \end{align*}$$

so $f(x)$ changes sign three times on the interval $[-4,4]$ and by the intermediate value theorem must have (at least) three zeroes.

Note that there are many ways to choose the test points, but using the critical points is an easy way to be sure of hitting good ones.

Added: To show that there are only three solutions to the original equation, apply Rolle’s theorem to $f$ to show that between any two zeroes of $f$ there must be a critical point.

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This looks good. I will try it from your angle. –  Yellow Skies Sep 27 '12 at 5:19
    
Oh yeah i was wondering how to prove that it has a maximum of 3 solutions? –  Yellow Skies Sep 27 '12 at 5:26
    
@TayBoonSiang: Rolle’s theorem: between any two solutions there must be a critical point of $f$. –  Brian M. Scott Sep 27 '12 at 5:27
    
Ok i need your help on this hehe. I am not sure how to use rolles theorem to show at modt 3 solution –  Yellow Skies Sep 27 '12 at 5:57
    
@TayBoonSiang: Rolle’s theorem says that if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, and if $f(a)=f(b)$, then there is a point $c\in(a,b)$ such that $f\,'(c)=0$. Think about what this means when $a$ and $b$ are zeroes of $f$: $f(a)=f(b)=0$, so there’s a critical point $c$ between $a$ and $b$. If this $f$ had four zeroes, $x_0<x_1<x_2<x_3$, it would have to have critical points $c_1,c_2,c_3$ such that $x_0<c_1<x_1<c_2<x_2<c_3<x_3$; is that possible for this function? –  Brian M. Scott Sep 27 '12 at 6:03

Continuity should do it. Evaluate $x^3-15x+1$ at enough places between $-4$ and $4$ to show that it changes sign three times.

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That is enough to show that it has at least three zeros, which strictly speaking may answer the question as stated. I wonder if it is intended that it be shown that there are no more than 3, in which case something more is needed (e.g., the derivative, or properties of zeros of polynomials). –  Jonas Meyer Sep 27 '12 at 5:09
    
Agreed on all points. Properties of zeros of polynomials would seem to be the easiest - how many zeros can a polynomial of degree 3 have? –  Gerry Myerson Sep 27 '12 at 5:10
    
Yeah i tried that. That is the exact question given. I'm not sure if they want me to make it exactly 3. Or they want me to show at least 3. –  Yellow Skies Sep 27 '12 at 5:18

You could use the intermediate value theorem. Write $f(x)=x^3 -15x+1$ and find pairs of positive and negative points on the interval. e.g. $f(0)=1$ and $f(1)=-13$. Use the intermediate value theorem to show that, since the curve is continuous, it must cross the x-axis. Continuing from the previous example, there exists an $x$ in $[0,1]$ s.t. $f(x)=0$.

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