Question

LEVEL-1
1. Find the zeros of each of the following quadratic polynomials and verify the
relationship between the zeros and their coefficients:
(i) f(x) = x2 - 2x - 8 (ii) g(s) = 4s2 -- 4s + 1

(3) h(t) = ? t2- 15 (iv) f(x) = 6x2 - 3 - 7x

(v) p(x) = x2 +212x - 6 (vi) q(x) = 13x2 + 10x + 713
please answer me fast!!​

Answer 1

Answer:

i) f(x) = x2 - 2x - 8

Condition 1

f(x) = x2 + 2x - 8 is continuous at (-4,2)

Since f(x) = x2 + 2x - 8 is a polynomial

& Every polynomial function is continuous for all x ER

= f(x) = x2 + 2x - 8 is continuous at x E [- 4, 2]f(x) = x2 - 2x - 8

Condition 2

f (x) = x² + 2x - 8 is differentiable at (-4, 2)

f(x) = x2 + 2x - 8 is a polynomial.

& Every polynomial function is differentiable for all x ER

therefore f(x) is differentiable at (-4,2)

Conditions of Rolle's theorem

1. f(x) is continuous at (a,b)

2. f(x) is derivable at (a,b)

3. f(a) = f(b) =

If all 3 conditions satisfied then there exist some c in (a, b) such that f'(c) = 0

Condition 3

f(x) = x2 + 2x - 8

f(-4)= (-4)2 + 2(-4) – 8 = 16 - 8 - 8 = 16 - 16 = 0

& f (2) = (2)2 + 2(2) – 8 = 4 + 4 - 8 = 8 - 8 = 0

Hence f(-4) = f(2) =

Now,

f(x) = x2 + 2x - 8

f'(x) = 2x + 2-0

f'(x) = 2x + 2

f'(x) = 2c + 2

Since all three conditions satisfied

f'(c) = 0

2c + 2 = 0

2c = - 2

c = -2/2 = - 1

Value of c = -1 E (-4,2)

Thus, Rolle's Theorem is satisfied.

Answer 2

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