If p(x) be a polynomial of degree three that has a local maximum value 8 at x = 1 and a lo…

Question

If p(x) be a polynomial of degree three that has
a local maximum value 8 at x = 1 and a local
minimum value 4 at x = 2; then p(0) is equal to :

Accepted Answer

Since p(x) has relative extreme at

x = 1 \& 2

so p'(x) = 0 at x = 1 \& 2

$$ \therefore $$ Let p'(x) = A(x -- 1) (x -- 2)

$$ \Rightarrow $$ p(x) = $$\int {A\left( {{x^2} - 3x + 2} \right)dx} $$

$$ \Rightarrow $$ p(x) = $${A\left( {{{{x^2}} \over 3} - 3{x^2} + 2x} \right)}$$ + C ...(1)

As P(1) = 8

From (1)

$$8 = A\left( {{1 \over 3} - {3 \over 2} + 2} \right) + C$$

$$ \Rightarrow $$ 8 = $${{5A} \over 6} + C$$

$$ \Rightarrow $$ 48 = 5A + 5C ...(2)

Also P(2) = 4

From (1)

$$4 = A\left( {{8 \over 3} - 6 + 4} \right) + C$$

$$ \Rightarrow $$ 4 = $${{2A} \over 3} + C$$

$$ \Rightarrow $$ 12 = 2A + 3C  ......(3)

Form (3) \& (4), C = --12, A = 24

Now p(0) = C = -12

Solutions · 1