Question

Let f(x) be a fourth degree polynomial with coefficient of x^4 is 1 such that f(-1) =1 , f(2) =4 , f(-3)=9 , f(4)=16 , then find f(1).

Answer 1

Let the polynomial f(x) = (x+1)(x-2)(x+3)(x-4) + x²
This polynomial satisfies all the given conditions.
To find f(1), let us plug x=1 into the equation
f(1) = 2×-1×4×-3 + 1 = 24+1 = 25

Answer 2

Answer:

The value of f(1) is 25.

Step-by-step explanation:

We are given that,

Polynomial f(x) is a fourth degree polynomial such that,

f(-1) = 1, f(2) = 4, f(-3) = 9, f(4) = 16.

Let us take the polynomial as,

$f(x)=(x+1)(x-2)(x+3)(x-4)+x^2$

So, we see that, this polynomial have degree 4.

Also, $f(-1)=(-1+1)(-1-2)(-1+3)(-1-4)+(-1)^2$ i.e. f(-1) = 1

$f(2)=(2+1)(2-2)(2+3)(2-4)+(2)^2$ i.e. f(2) = 4

$f(-3)=(-3+1)(-3-2)(-3+3)(-3-4)+(-3)^2$ i.e. f(-3) = 9

$f(4)=(4+1)(4-2)(4+3)(4-4)+(4)^2$ i.e. f(4) = 16

So, we need to find the value of f(1).

On substituting, we get,

$f(1)=(1+1)(1-2)(1+3)(1-4)+(1)^2$

i.e. $f(1)=2\times (-1)\times 4\times (-3)+1$

i.e. $f(1)=24+1$

i.e. $f(1)=25$

Thus, the value of f(1) is 25.