UPSKILL MATH PLUS
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Learn moreAnswer variants:
\(\frac{2}{2x^2 - 1}\)
\(4x^2\)
\(4x^2 + 8x + 3\)
\(\frac{8}{x^2} - 1\)
\(x^2 - 12x + 36\)
\(\frac{3 - x}{3} - 1\)
\(\frac{9 - x}{3}\)
\(x - 1\)
\(x^2 - 6\)
Using the functions \(f\) and \(g\) given below, find \(f \circ g\) and \(g \circ f\). Check whether \(f \circ g = g \circ f\).
(i) \(f(x) = x - 6\), \(g(x) = x^2\)
\(f \circ g\) \(=\)
\(g \circ f\) \(=\)
So,
(ii) \(f(x) = \frac{2}{x}\), \(g(x) = 2x^2 - 1\)
\(f \circ g\) \(=\)
\(g \circ f\) \(=\)
So, .
(iii) \(f(x) = \frac{x + 6}{3}\), \(g(x) = 3 - x\)
\(f \circ g\) \(=\)
\(g \circ f\) \(=\)
So, .
(iv) \(f(x) = 3 + x\), \(g(x) = x - 4\)
\(f \circ g\) \(=\)
\(g \circ f\) \(=\)
So, .
(v) \(f(x) = 4x^2 - 1\), \(g(x) = 1 + x\)
\(f \circ g\) \(=\)
\(g \circ f\) \(=\)
So, .
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