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If ๐‘“(๐‘ฅ)=๐‘ฅ2+3 and ๐‘”(๐‘ฅ)=7๐‘ฅโˆ’3, find ๐‘“(๐‘”(๐‘ฅ)).๐‘“(๐‘”(๐‘ฅ))

Question

If ๐‘“(๐‘ฅ)=๐‘ฅ2+3 and ๐‘”(๐‘ฅ)=7๐‘ฅโˆ’3, find ๐‘“(๐‘”(๐‘ฅ)).๐‘“(๐‘”(๐‘ฅ))

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Solution

To find the composition of the functions ๐‘“(๐‘”(๐‘ฅ)), we need to substitute ๐‘”(๐‘ฅ) into ๐‘“(๐‘ฅ).

Step 1: Write down the function ๐‘“(๐‘ฅ)=๐‘ฅยฒ+3

Step 2: Substitute ๐‘”(๐‘ฅ) into ๐‘“(๐‘ฅ). So wherever there is an 'x' in ๐‘“(๐‘ฅ), we replace it with ๐‘”(๐‘ฅ).

This gives us ๐‘“(๐‘”(๐‘ฅ)) = (7๐‘ฅ-3)ยฒ + 3

Step 3: Expand the square in the equation.

(7๐‘ฅ-3)ยฒ = 49๐‘ฅยฒ - 42๐‘ฅ + 9

So, ๐‘“(๐‘”(๐‘ฅ)) = 49๐‘ฅยฒ - 42๐‘ฅ + 9 + 3

Step 4: Simplify the equation by adding the constants.

So, ๐‘“(๐‘”(๐‘ฅ)) = 49๐‘ฅยฒ - 42๐‘ฅ + 12

So the composition of the functions ๐‘“(๐‘”(๐‘ฅ)) is 49๐‘ฅยฒ - 42๐‘ฅ + 12.

This problem has been solved

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