Questions: f(n)=2n+4 g(n)=n^3-4n^2 Find f(1)-g(1)

f(n)=2n+4 g(n)=n^3-4n^2 Find f(1)-g(1)
Transcript text: 7) \[ \begin{array}{l} f(n)=2 n+4 \\ g(n)=n^{3}-4 n^{2} \\ \text { Find } f(1)-g(1) \end{array} \]
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Solution

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Solution Steps

To solve the problem, we need to evaluate the functions \( f(n) \) and \( g(n) \) at \( n = 1 \). Then, we subtract the result of \( g(1) \) from \( f(1) \).

Step 1: Evaluate \( f(1) \)

The function \( f(n) \) is given by: \[ f(n) = 2n + 4 \] Substituting \( n = 1 \): \[ f(1) = 2(1) + 4 = 2 + 4 = 6 \]

Step 2: Evaluate \( g(1) \)

The function \( g(n) \) is given by: \[ g(n) = n^3 - 4n^2 \] Substituting \( n = 1 \): \[ g(1) = 1^3 - 4(1)^2 = 1 - 4 = -3 \]

Step 3: Calculate \( f(1) - g(1) \)

Now, we find the difference: \[ f(1) - g(1) = 6 - (-3) = 6 + 3 = 9 \]

Final Answer

\(\boxed{9}\)

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