Question

find the value of each of the following (i) 3y^3-4y+ $\sqrt{} } 11$ at y=2 (ii) 4x^3-7x+8 at x=2

Answer 1

1.  3y³-4y+ √11  at  y=2  is  16+√11

2. 4x³-7x+8   at  x=2  is 26

Explanation:

Given:

1. 3y³-4y+ √11 at y=2

2. 4x³-7x+8  at x=2

To find:

The value of x and y

Solution:

1. 3y³-4y+ √11 at y=2

==> p(y) =3y³-4y+ √11

==> p(2) = 3(2)³-4(2)+√11

==> p(2) = 3(8)-4(2)+√11

==> p(2) = 3(8)-8+√11

==> p(2) = 24-8+√11

==> p(2) = 16+√11

==> 3y³-4y+ √11 at y=2 is  16+√11

2. 4x³-7x+8  at x=2

==> p(x) =4x³-7x+8

==> p(2) = 4(2)³-7(2)+8

==> p(2) = 4(8)-7(2)+8

==> p(2) = 32-14+8

==> p(2) = 18+8

==> p(2) = 26

==> 4x³-7x+8  at x=2 is 26