Exercise:
Let \(p(x)\) be a polynomial of 2nd degree which satisfies \( p(2) =10 , p'(2) =12\) & \( p''(2)=6\)
1) Determine \(p(x)\)
2) Solve for \(p(x)=\sqrt{3}x\)
Solution:
1) Consider \(p(x)\) the polynomial of 2nd degree thus \(p(x)=a{x}^{2}+bx+c\) so for \(p(2)=10\) then \(10=4a+2b+c...(1) \) also we have \(p'(2)=12\) and we know that \(p'(x)=2ax+b\) and \(p''(x)=2a\) thus \(12=2a(2)+b ....(2)\) and \(6=2a\) thus \(a=\frac{6}{2}=3 \) now back to (2) to get \(b\) i.e \(12=4(3)+b\)
thus \(b=0\) from (1) we get \(10=12+0+c\) thus \(c=-2\)
hence \(p(x)=3{x}^{2}-2\)
2) we have \(p(x)=3{x}^{2}-2=\sqrt{3}x\) thus \(3{x}^{2}-\sqrt{3}x-2=0 \) so
Let \(p(x)\) be a polynomial of 2nd degree which satisfies \( p(2) =10 , p'(2) =12\) & \( p''(2)=6\)
1) Determine \(p(x)\)
2) Solve for \(p(x)=\sqrt{3}x\)
Solution:
1) Consider \(p(x)\) the polynomial of 2nd degree thus \(p(x)=a{x}^{2}+bx+c\) so for \(p(2)=10\) then \(10=4a+2b+c...(1) \) also we have \(p'(2)=12\) and we know that \(p'(x)=2ax+b\) and \(p''(x)=2a\) thus \(12=2a(2)+b ....(2)\) and \(6=2a\) thus \(a=\frac{6}{2}=3 \) now back to (2) to get \(b\) i.e \(12=4(3)+b\)
thus \(b=0\) from (1) we get \(10=12+0+c\) thus \(c=-2\)
hence \(p(x)=3{x}^{2}-2\)
2) we have \(p(x)=3{x}^{2}-2=\sqrt{3}x\) thus \(3{x}^{2}-\sqrt{3}x-2=0 \) so
you wrote $P(2) = 12$, but it should be $P(2)=10$
ReplyDeletethank you , you are right
Delete