Question

If ƒ(x) is polynomial of degree 4 such that ƒ(1) = 1, ƒ(2) = 2, ƒ(3) = 3, ƒ(4) = 4 & ƒ(0) = 1

find ƒ(5).​

Answer 1

Hint

As

• $f(1)-1=0$
• $f(2)-2=0$
• $f(3)-3=0$
• $f(4)-4=0$

we apply factor theorem on the polynomial $f(x)-x$. It has $x=1,2,3,4$ as its zeros.

Solution

Using factor theorem on $f(x)-x$, we see that $f(x)-x=a(x-1)(x-2)(x-3)(x-4)$ is the polynomial of degree 4.

The next to do is to find the coefficient of the leading term.

When $x=0$

$\implies f(0)-0=a\cdot (-1)\cdot (-2)\cdot (-3)\cdot (-4)$

So

$\implies 24a=1$

$\implies a=\dfrac{1}{24}$

The equation of degree 4 is $f(x)=\dfrac{1}{24} (x-1)(x-2)(x-3)(x-4)+x$.

When $x=5$

$\implies f(5)=\dfrac{1}{24} \cdot (4)\cdot (3)\cdot (2)\cdot (1)+5$

$\implies \boxed{f(5)=6}$

Answer 2

Given :-

If ƒ(x) is polynomial of degree 4 such that ƒ(1) = 1, ƒ(2) = 2, ƒ(3) = 3, ƒ(4) = 4 & ƒ(0) = 1

To Find :-

ƒ(5).​

Solution :-

x = 1

x - 1 = 0

x = 2

x - 2 = 0

x = 3

x - 3 = 0

x = 4

x - 4 = 0

Now

When x = 0

f(0) = a × (0 - 1) × (0 - 2) × (0 - 3) × (0 - 4)

1 = a × -1 × -2 × -3 × -4

1 = 24a

1/24 = a

Now

By putting as 1/24

f(x) = 1/24 × (x - 1) × (x - 2) × (x - 3) × (x - 4) + x

Let x = 5

f(5) = 1/24 × (5 - 1) × (5 - 2) × (5 - 3) × (5 - 4) × 5

f(5) = 1/24 × 4 × 3 × 2 × 1 × 5

f(5) = 1/24 × 12 × 2 × 5

f(5) = 1/24 × 120

f(5) = 1 × 5

f(5) = 5