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Show that the equation $ x^3 - 15x + c = 0 $ has at most one root in the interval $ [-2, 2] $.

Let $f(x)=x^{3}-15 x+c$ for $x$ in $[-2,2] .$ If $f$ has two real roots $a$ and $b$ in $[-2,2],$ with $a<b,$ then $f(a)=f(b)=0 .$ since

the polynomial $f$ is continuous on $[a, b]$ and differentiable on $(a, b),$ Rolle's Theorem implies that there is a number $r$ in $(a, b)$

such that $f^{\prime}(r)=0 .$ Now $f^{\prime}(r)=3 r^{2}-15 .$ since $r$ is in $(a, b),$ which is contained in $[-2,2],$ we have $|r|<2,$ so $r^{2}<4$

It follows that $3 r^{2}-15<3 \cdot 4-15=-3<0 .$ This contradicts $f^{\prime}(r)=0,$ so the given equation can't have two real roots

in $[-2,2] .$ Hence, it has at most one real root in [-2,2]

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explicitly worried here. So you have to prove that I can't cross the X axis twice on negative to Hama tubes and which proves that it has a most one route. So we have ever bats is equal to X cubed minus 15 X plus C. So it's continuous on negative, too common to, since it's a polynomial and it's also differential. Since the derivative is equal to three X Square minus 15 we're gonna see if it's increasing or decreasing or both in the interval. Negative to comment, too. So these are derivative and we plug in negative two and two. So negative two gives us negative three 12 gives us negative three. These are both less than zero, Which means Dr F is on Lee decreasing which states that could cross the ex access only once. Then it proves that has, at most one brute. And then we're just gonna plug these back our values and we get negative too on two. So we have at most one Ruth