if one root of the polynomial f(x)=3^3+11x+p is reciprocal of the other then find value of p
Answers
Step-by-step explanation:
Since \cos(x)cos(x) is a periodic function, the inverse function \cos^{-1}(x)cos
−1
(x) is only defined for -1\leq x\leq1−1≤x≤1 .
So, let's find the value in the domain by using the properties of \cos(x)cos(x) .
\hookrightarrow\cos(\dfrac{7\pi}{6})=\cos(2\pi-\dfrac{7\pi}{6})↪cos(
6
7π
)=cos(2π−
6
7π
)
\hookrightarrow\cos(\dfrac{7\pi}{6})=\cos(\dfrac{5\pi}{6})↪cos(
6
7π
)=cos(
6
5π
)
We know that the inverse function is a kind of function that assigns a domain value for the range.
So,
\hookrightarrow f^{-1}(f(x))=x↪f
−1
(f(x))=x
So,
\large\hookrightarrow\red{\boxed{\red{\bold{\cos^{-1}(\cos(\dfrac{7\pi}{6}))=\red{\underline{\dfrac{5\pi}{6}}}}}}}↪
cos
−1
(cos(
6
7π
))=
6
5π
This is our answer.
\arcsin(x),\ \arccos(x),\ \arctan(x)arcsin(x), arccos(x), arctan(x) refers to \sin^{-1}(x),\ \cos^{-1}(x),\ \tan^{-1}(x)sin
−1
(x), cos
−1
(x), tan
−1
(x) .
For the restricted range of \sin(x),\ \cos(x),\ \tan(x)sin(x), cos(x), tan(x) , view \text{[Attachment 1, 2, 3]}[Attachment 1, 2, 3] .
The domains of \arcsin(x)arcsin(x) and \arccos(x)arccos(x) are,
\large\hookrightarrow\red{\boxed{\red{\bold{-1\leq x\leq1}}}}↪
−1≤x≤1