Click and drag the steps to their corresponding step numbers to prove that the given pair of functions are of the same order. (Note: Consider to prove the result, first prove (x) = O(g(x)) and then prove g(x) = O({X)). AX) = 2x2 + x - 7 and g(x) = x- Step 1 For large x, 2x² + x 7 2 3x2. Hence |f(x)| 2 1g(x) for large x. For large x, 2x² + x 7 s 3x². Hence |f(x)| < 3g(x) for large x. For large x, x2 > 2x2 + x – 7. Hence |f(x)| 2 3g(x) for large x. Step 2 For large x, x² S 2x² + x – 7. Hence, Ig(x)| S 1 |f(x)| for large x. For large x, x2 s 2x² + X – 7. Hence, Įg(x)| S 1 |f(x)| for large x. Step 3 Hence, f(x) = O(g(x)) and g(x) = O(A{x)). Hence, f(x) = O(g(x)) and g(x) = O(f\x)). For large x, 2x² + x 7 s 3x². Hence |f(x)| < 3g(x) for large x.
Click and drag the steps to their corresponding step numbers to prove that the given pair of functions are of the same order. (Note: Consider to prove the result, first prove (x) = O(g(x)) and then prove g(x) = O({X)). AX) = 2x2 + x - 7 and g(x) = x- Step 1 For large x, 2x² + x 7 2 3x2. Hence |f(x)| 2 1g(x) for large x. For large x, 2x² + x 7 s 3x². Hence |f(x)| < 3g(x) for large x. For large x, x2 > 2x2 + x – 7. Hence |f(x)| 2 3g(x) for large x. Step 2 For large x, x² S 2x² + x – 7. Hence, Ig(x)| S 1 |f(x)| for large x. For large x, x2 s 2x² + X – 7. Hence, Įg(x)| S 1 |f(x)| for large x. Step 3 Hence, f(x) = O(g(x)) and g(x) = O(A{x)). Hence, f(x) = O(g(x)) and g(x) = O(f\x)). For large x, 2x² + x 7 s 3x². Hence |f(x)| < 3g(x) for large x.
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter3: Functions
Section3.3: More On Functions; Piecewise-defined Functions
Problem 98E: Determine if the statemment is true or false. If the statement is false, then correct it and make it...
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Click and drag the steps to their corresponding step numbers to prove that the given pair of functions are of the same order.
(Note: Consider to prove the result, first prove f(x) = O(g(x)) and then prove g(x) = O(f(x)).
f(x) = 2x2 + x - 7 and g(x) = x2
Transcribed Image Text:Click and drag the steps to their corresponding step numbers to prove that the given pair of functions are of the same
order.
(Note: Consider to prove the result, first prove fx) = O(g(x)) and then prove g(x) = 0(Ax).
Ax) =
) = 2x2
+ x - 7 and g(x) = x²
Step 1
For large x, 2x² + x 7 2 3x².
Hence |f(x)| 2 1g(x) for large x.
For large x, 2x² + x 7< 3x?.
Hence |f(x)| s 3g(x) for large x.
For large x, x2z 2x2 + x – 7.
Hence |f(x)| 2 3g(x) for large x.
Step 2
For large x, x?S 2x² + x – 7.
Hence, Įg(x)| s 1 |f(X)| for large x.
For large x, x² s 2x² + X – 7.
Hence, [g(x)| < 1 |f(x)| for large x.
Step 3
Hence, f(x) = O(g(x)) and g(x) =
O(f(x)).
Hence, f(x) = O(g(x)) and g(x) =
O(f(x)).
For large x, 2x²+ x 7< 3x².
Hence |f(x)| < 3g(x) for large x.
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