A polynomial function f(x) of degree 5 with leading coefficients one, increase in the interval (-infty, 1) and (3,infty) and decrease in the interval (1, 3). Given that f′(2)=0,f(0)=4. Then value of f′(6)=?

Ivan Buckley

Ivan Buckley

Answered question

2022-09-25

Value of f′(6) in given polynomial
A polynomial function f(x) of degree 5 with leading coefficients one , increase in the interval
( , 1 ) and ( 3 , ) and decrease in the interval (1,3). Given that f ( 2 ) = 0 , f ( 0 ) = 4.
Then value of f ( 6 ) =
Attempt: Given function increase in the interval ( , 1 ) and ( 3 , ) and decrease
in the interval (1,3). So we have f ( 1 ) = 0 and f ( 3 ) = 0.
So f ( x ) = ( x 1 ) ( x 2 ) ( x 3 ) ( x α ).
Could some help me how i calculate α ,

Answer & Explanation

Cassie Moody

Cassie Moody

Beginner2022-09-26Added 10 answers

Step 1
Since f decreases on the entire interval (1,3), the root of f′ at 2 must have even multiplicity. It cannot be 4-fold or more (since that would make the degree of f′ as a polynomial too high), so it must be a double root, and your α = 2.
Step 2
Note also that the leading coefficient of f′ is 5, which you had forgotten in your development.
Valentina Holland

Valentina Holland

Beginner2022-09-27Added 5 answers

Step 1
f ( x ) = x 5 + a x 4 + b x 3 + c x 2 + d x + 4
f ( 2 ) = 80 + 32 a + 12 b + 4 c + d = 0
f ( 1 ) = 5 + 4 a + 3 b + 2 c + d = 0
f ( 3 ) = 405 + 108 a + 27 b + 6 c + d = 0
f ( 6 ) = 5.6 4 + 4.6 3 . a + 3.6 2 b + 12 c + d

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