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When ๐‘“(๐‘ฅ) = 3๐‘ฅ โˆ’ 5 and ๐‘”(๐‘ฅ) = 2๐‘ฅ2 โˆ’ 5 , find ๐‘“(๐‘”(๐‘ฅ))

Question

When ๐‘“(๐‘ฅ) = 3๐‘ฅ โˆ’ 5 and ๐‘”(๐‘ฅ) = 2๐‘ฅ2 โˆ’ 5 , find ๐‘“(๐‘”(๐‘ฅ))

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Solution

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

Step 1: Write down the function ๐‘“(๐‘ฅ) = 3๐‘ฅ โˆ’ 5

Step 2: Substitute ๐‘”(๐‘ฅ) = 2๐‘ฅยฒ - 5 into ๐‘“(๐‘ฅ). Wherever there is an ๐‘ฅ in the function ๐‘“, replace it with ๐‘”(๐‘ฅ).

So, ๐‘“(๐‘”(๐‘ฅ)) = 3(๐‘”(๐‘ฅ)) - 5 = 3(2๐‘ฅยฒ - 5) - 5

Step 3: Simplify the expression.

๐‘“(๐‘”(๐‘ฅ)) = 6๐‘ฅยฒ - 15 - 5 = 6๐‘ฅยฒ - 20

So, ๐‘“(๐‘”(๐‘ฅ)) = 6๐‘ฅยฒ - 20.

This problem has been solved

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