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प्रश्न
For the curve y = `x^((1/3))` given in Figure 1.68, draw
y = `x^((1/3)) - 1`
उत्तर
y = `x^((1/3)) - 1`
y + 1 = `x^((1/3))`
(y + 1)3 = x
When y = 0 ⇒ (0 + 1)3 = x ⇒ x = 1
y = 1 ⇒ (1 + 1)3 = x ⇒ x = 8
y = 2 ⇒ (2 + 1)3 = x ⇒ x = 27
y = – 1 ⇒ (– 1 + 1)3 = x ⇒ x = 0
y = – 2 ⇒ (– 2 + 1)3 = x ⇒ x = – 1
y = – 3 ⇒ (– 3 + 1)3 = x ⇒ x = – 8
| x | 1 | 8 | 27 | 0 | – 1 | – 8 |
| y | 0 | 1 | 2 | – 1 | – 2 | – 3 |
The graph of y = `x^((1/3)) - 1` causes the graph y = `x^((1/3))` a shift to the downward by 1 unit.
The graph of y = f(x) – d, d > 0 causes the graph y = f(x) a shift to the downward by d units.
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संबंधित प्रश्न
For the curve y = x3 given in Figure 1.67, draw
y = −x3 
For the curve y = x3 given in Figure 1.67, draw
y = x3 + 1
For the curve y = x3 given in Figure 1.67, draw
y = x3 − 1
For the curve y = x3 given in Figure 1.67, draw
y = (x + 1)3 with the same scale
For the curve y = `x^((1/3))` given in Figure 1.68, draw
y = `- x^((1/3))`
For the curve y = `x^((1/3))` given in Figure 1.68, draw
y = `x^((1/3)) + 1`
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y = sin(− x)
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y = − sin(−x)
From the curve y = x, draw y = − x
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From the curve y = x, draw y = x + 1
From the curve y = x, draw 2x + y + 3 = 0
From the curve y = |x|, draw y = |x − 1| + 1
From the curve y = |x|, draw y = |x + 2| − 3
From the curve y = sin x, draw y = sin |x| (Hint: sin(−x) = − sin x)
