If #f(x) = x^2 - x# and #g(x) = 3x + 1# how do you find #g(f(sqrt2))#?

Answer 1

To find ( g(f(\sqrt{2})) ), first, find ( f(\sqrt{2}) ) by substituting ( \sqrt{2} ) into the function ( f(x) = x^2 - x ). Then, take the result and substitute it into the function ( g(x) = 3x + 1 ) to find the final answer.

Find ( f(\sqrt{2}) ): [ f(\sqrt{2}) = (\sqrt{2})^2 - \sqrt{2} = 2 - \sqrt{2} ]
Substitute ( f(\sqrt{2}) = 2 - \sqrt{2} ) into ( g(x) = 3x + 1 ): [ g(2 - \sqrt{2}) = 3(2 - \sqrt{2}) + 1 = 6 - 3\sqrt{2} + 1 = 7 - 3\sqrt{2} ]

So, ( g(f(\sqrt{2})) = 7 - 3\sqrt{2} ).

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Answer 2

Abigail Allen

#7-3sqrt2#

#f(color(red)(sqrt2))=(color(red)(sqrt2))^2-color(red)(sqrt2)=2-sqrt2#

#"to evaluate " g(f(sqrt2)))#

#"substitute " f(sqrt2)" into " g(x)#

#rArrg(f(sqrt2)))=3(2-sqrt2)+1#

#color(white)(rArrg(f(sqrt2x)))=6-3sqrt2+1#

#color(white)(rArrg(f(sqrt2x)))=7-3sqrt2#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.