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junior college 1 | H2 Maths
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Date Posted: 4 years ago
Views: 343
What is f(x) ?
F(x)= λ + (1/1-x), x∈R, x ≠1
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)
= ( 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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