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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 = 0
y = 2 ⇒ (2 – 1)3 = x ⇒ x = 1
y = 3 ⇒ (3 – 1)3 = x ⇒ x = 8
y = – 1 ⇒ (– 1 – 1)3 = x ⇒ x = – 8
y = – 2 ⇒ (– 2 – 1)3 = x ⇒ x = – 27
| x | – 1 | 0 | 1 | 8 | – 8 | – 27 |
| y | 0 | 1 | 2 | 3 | – 1 | – 2 |
The graph of y = `x^((1/3)) + 1` causes the graph y = `x^((1/3))` a shift to the upward by 1 unit.
The graph of y = f(x) + d, d > 0 causes the graph y = f(x) a shift to the upward by d units.
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संबंधित प्रश्न
For the curve y = x3 given in Figure 1.67, draw
y = x3 − 1 
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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From the curve y = sin x, graph the function
y = `sin (pi/2 - x)` which is also cos x (refer trigonometry)
From the curve y = x, draw y = − x
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From the curve y = x, draw y = x + 1
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From the curve y = |x|, draw y = |x + 1| − 1
