A polynomial function P ( x ) of degree 5 with leading coefficient one, increases in the interval (-∞, 1) and (3, ∞) and decreases in the interval (1,3). Given that P(0)=4 and P prime(2)=0. Find the value P prime(6).

Question

A polynomial function P ( x ) of degree 5 with leading coefficient one, increases in the interval (-∞, 1) and (3, ∞) and decreases in the interval (1,3). Given that P(0)=4 and P prime(2)=0. Find the value P prime(6). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

∵ Degree of P ( x ) is 5 with leading coefficient one. ∴ text degree of P prime( x ) text is 4 text with leading coefficient five. ∴ P prime( x )=5( x -1)( x -3)( x -2)2 ∴ P prime(6)=5 × 5 × 3 × 42 =1200

Q. A polynomial function $P (x)$ of degree 5 with leading coefficient one, increases in the interval $(- \infty, 1)$ and $(3, \infty)$ and decreases in the interval $(1, 3)$ . Given that $P (0) = 4$ and $P^{'} (2) = 0$ . Find the value $P^{'} (6)$ .

$∵$ Degree of $P (x)$ is 5 with leading coefficient one.
$∴ degree of P^{'} (x) is 4 with leading coefficient five.$
$∴ P^{'} (x) = 5 (x - 1) (x - 3) (x - 2)^{2}$
$∴ P^{'} (6) = 5 \times 5 \times 3 \times 4^{2}$
$= 1200$

Q. A polynomial function P(x) of degree 5 with leading coefficient one, increases in the interval (−∞,1) and (3,∞) and decreases in the interval (1,3). Given that P(0)=4 and P′(2)=0. Find the value P′(6).

∵ Degree of P(x) is 5 with leading coefficient one. ∴ degree of P′(x) is 4 with leading coefficient five. ∴P′(x)=5(x−1)(x−3)(x−2)2 ∴P′(6)=5×5×3×42 =1200

Q. A polynomial function $P (x)$ of degree 5 with leading coefficient one, increases in the interval $(- \infty, 1)$ and $(3, \infty)$ and decreases in the interval $(1, 3)$ . Given that $P (0) = 4$ and $P^{'} (2) = 0$ . Find the value $P^{'} (6)$ .

$∵$ Degree of $P (x)$ is 5 with leading coefficient one.
$∴ degree of P^{'} (x) is 4 with leading coefficient five.$
$∴ P^{'} (x) = 5 (x - 1) (x - 3) (x - 2)^{2}$
$∴ P^{'} (6) = 5 \times 5 \times 3 \times 4^{2}$
$= 1200$