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प्रश्न
For the curve y = `x^((1/3))` given in Figure 1.68, draw
y = `(x + 1)^((1/3))`
उत्तर
y = `(x + 1)^((1/3))`
y3 = x + 1
When y = 0 ⇒ 03 = x + 1 ⇒ x = – 1
y = 1 ⇒ 13 = x + 1 ⇒ x = 0
y = 2 ⇒ 23 = x + 1 ⇒ x = 8 – 1 = 7
y = 3 ⇒ 33 = x + 1 ⇒ x = 27 – 1 = 26
y = – 1 ⇒ (– 1)3 = x + 1 ⇒ x = – 1 – 1 = – 2
y = – 2 ⇒ (– 2)3 = x + 1 ⇒ x = – 8 – 1 = – 9
| x | – 1 | 0 | 7 | 26 | – 2 | – 9 |
| y | 0 | 1 | 2 | 3 | – 1 | – 2 |
The graph of y = (x + 1)3 causes the graph y = `x^((1/3))` a shift to the left by 1 unit.
The graph of y = f(x + c), c > 0 causes the graph y = f(x) a shift to the left by c 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 = `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`
For the curve y = `x^((1/3))` given in Figure 1.68, draw
y = `x^((1/3)) - 1`
Graph the functions f(x) = x3 and g(x) = `root(3)(x)` on the same coordinate plane. Find f o g and graph it on the plane as well. Explain your results
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From the curve y = sin x, graph the function.
y = sin(− x)
From the curve y = sin x, graph the function
y = − sin(−x)
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
From the curve y = x, draw y = x + 1
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)
