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junior college 1 | H2 Maths
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Kayre
Kayre

junior college 1 chevron_right H2 Maths chevron_right Singapore

need help thx,

Date Posted: 3 years ago
Views: 313
J
J
3 years ago
What is f(x) ?
Kayre
Kayre
3 years ago
F(x)= λ + (1/1-x), x∈R, x ≠1
J
J
3 years ago
hg(x)

= ( λ + 1/(1-x) - (λ + ⅔) )²

= (1/(1- x) - ⅔)²

Dhg = Dg = (-∞, 0)

Now as x →-∞, -x→∞

So 1 - x →∞
1/(1 - x) → 0
1/(1-x) - ⅔ → -⅔
(1/(1-x) - ⅔)² →(-⅔)² = 4/9

When x = 0,

(1/(1-x) - ⅔)² = (1/1 - ⅔)²
= (⅓)²
= 1/9

So this is not bigger than 4/9.


Since hg(x) has a squared, it is either 0 or positive for all real values of x
i.e hg(x) ≥ 0 for x ∈ R

The smallest value of hg(x) = 0

So Rhg = (0,4/9)


Extra : to find the minimum point

(1/(1-x) - ⅔)² = 0

1/(1-x) - ⅔ = 0

1/(1-x) = ⅔

1 - x = 3/2

x = -½

The minimum point is at (-½,0)

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No f(x) how to solve?
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Kahwai
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