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mariarad [96]
3 years ago
14

The question is:

Mathematics
1 answer:
saul85 [17]3 years ago
8 0
f(x) = \sqrt{x+1} \\ y = \sqrt{x+1} \ or \ y^2 = x+1 \ or \ x = y^2 - 1 \\ Therefore, \ f^{-1}(x)=x^2 - 1

(fof^{-1})(a) = f(f^{-1}(a))=f(a^2-1) \\ =\sqrt{a^2-1+1}=\sqrt{a^2}=a \\  \\ (f^{-1}of)(a)=f^{-1}(f(a))=f^{-1}(\sqrt{a+1})= \\ (\sqrt{a+1})^2-1 = a+1-1 = a

Therefore, (fof^{-1})(a) = (f^{-1}of)(a) = a

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An exponential growth function has an asymptote of y = –3. Which might have occurred in the original function to permit the rang
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The right choice is: A <em>whole</em> number constant could have been <em>subtracted</em> from the <em>exponential</em> expression.

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To learn more on asymptotes, we kindly invite to check this verified question: brainly.com/question/8493280

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The exponent tells us the number of times the decimal has to move. The sign of the exponent tells us the direction of the movement of the decimal.

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