- #1
LucasCLarson
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show x^3-15x+c=0 has at most one root on the interval [-2, 2].
My work/attempted solution.
Using the IVT i determined that there is a possible root (N=c) within the interval of [-2,2]. because c is my constant i decided to Use [d] as my letter for explanation.
f(x)=x^3-15x+c
f(d)=d^3-15d+c
0=d^3-15d+c
-c=d^3-15d
f(-2)=22=-c
c=-22 when for f(-2)
f(2)=-22=-c
c=22 for f(2)
f is continuous on [-2,2] because it is a polynomial. Let -c be a number b/w f(-2) and f(2) where f(-2) does not equal f(2). Then there exist a root d on (-2,2) such that F(d)=-c
I am stuck there. I feel like i have done a bunch of stuff wrong so a little nudge in the right direction would be great.
Thanks
My work/attempted solution.
Using the IVT i determined that there is a possible root (N=c) within the interval of [-2,2]. because c is my constant i decided to Use [d] as my letter for explanation.
f(x)=x^3-15x+c
f(d)=d^3-15d+c
0=d^3-15d+c
-c=d^3-15d
f(-2)=22=-c
c=-22 when for f(-2)
f(2)=-22=-c
c=22 for f(2)
f is continuous on [-2,2] because it is a polynomial. Let -c be a number b/w f(-2) and f(2) where f(-2) does not equal f(2). Then there exist a root d on (-2,2) such that F(d)=-c
I am stuck there. I feel like i have done a bunch of stuff wrong so a little nudge in the right direction would be great.
Thanks