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प्रश्न
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
उत्तर
Given functions are f(x) = x3 and g(x) = `x^((1/3))`
fog(x) = f(g(x))
= `f(x^(1/3))`
= `(x^(1/3))^3` = x
f(x) = x3
| x | 0 | 1 | – 1 | 2 | – 2 |
| y | 0 | 1 | – 1 | 8 | – 8 |
g(x) = `x^((1/3))`
| x | 0 | 1 | – 1 | 8 | – 8 |
| y | 0 | 1 | – 1 | 2 | – 2 |
Graph of fog(x) = x
| x | 0 | 1 | – 1 | 2 | – 2 | 3 |
| y | 0 | 1 | – 1 | 2 | – 2 | 3 |
Since fog(x) = x is symmetric about the line y = x, g(x) is the inverse image of f(x).
∴ g(x) = f–1(x)
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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`
For the curve y = `x^((1/3))` given in Figure 1.68, draw
y = `(x + 1)^((1/3))`
Write the steps to obtain the graph of the function y = 3(x − 1)2 + 5 from the graph y = x2
From the curve y = sin x, graph the function.
y = sin(− x)
From the curve y = sin x, graph the function
y = − sin(−x)
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y = `sin(pi/2 + x)` which is cos 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 = 2x
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)
