If f - R - R is defined by fx-2x-2- then f- f x - 2 -

The correct option is A 2f(x) Let f : R → R #160;is defined by #160; f(x)=2x 2 . #160;Then #160; (f∘ f)(x)+2= f[f(x)]+2 =f(2x ...

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How to Find the Inverse of a Function

If f : R → ...

Question

If $f : R \to R$ is defined by $f (x) = 2 x - 2,$ then $(f \circ f) (x) + 2 =$

f (x)

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2 f (x)

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3 f (x)

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- f (x)

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Solution

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Similar questions

f : R \to R

f (x) = 2 x + | x |

f (2 x) + f (- x) - f (x) = 2 | x |

f (x) = \frac{x + 2}{2 x + 3}

\int {(\frac{f (x)}{x^{2}})}^{1 / 2} d x = \frac{1}{\sqrt{2}} g (\frac{1 + \sqrt{2 f (x)}}{1 - \sqrt{2 f (x)}}) - \sqrt{\frac{2}{3}} h (\frac{\sqrt{3 f (x)} + \sqrt{2}}{\sqrt{3 f (x)} - \sqrt{2}}) + c

f : R \to R

f (x) = 2 x + 5

x > 0; f (x) = 3 x - 2

x \leq 0 :

f

f (x) = \frac{x + 2}{2 x + 3}

\int (\frac{f (x)}{x^{2}})^{1 / 2} d x

\frac{1}{\sqrt{2}} g (\frac{1 + \sqrt{2 f (x)}}{1 - \sqrt{2 f (x)}}) - \sqrt{\frac{2}{3}} h (\frac{\sqrt{3 f (x)} + \sqrt{2}}{\sqrt{3 f (x)} - \sqrt{2}}) + c

f : R \to R

f (x) = 3 x - 2

f (x) + 2

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