Question

investigate whether g(x) = 5+ e^2x+In6 is a possible primitive function to g(x) = 3e^2x+In4 Justify your answer.

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1. Compute the derivative of \( g(x) = 5 + e^{2x} + \ln{6} \):

g'(x) = \frac{d}{dx} \left( 5 + e^{2x} + \ln{6} \right)

2. Use the constant rule and the chain rule for differentiation:

\frac{d}{dx}(5) = 0

\frac{d}{dx}(e^{2x}) = 2e^{2x}

\frac{d}{dx}(\ln{6}) = 0

3. Combine the results:

g'(x) = 0 + 2e^{2x} + 0

g'(x) = 2e^{2x}

4. Compare \( g'(x) \) with \( h(x) \):

2e^{2x} \neq 3e^{2x} + \ln{4}

Since \( g'(x) \neq h(x) \), the function \( g(x) = 5 + e^{2x} + \ln{6} \) is not a primitive function of \( h(x) = 3e^{2x} + \ln{4} \).

Therefore, the answer is:

\boxed{\text{No, } g(x) = 5 + e^{2x} + \ln{6} \text{ is not a primitive function of } h(x) = 3e^{2x} + \ln{4}}

2. Use the constant rule and the chain rule for differentiation:

3. Combine the results:

4. Compare \( g'(x) \) with \( h(x) \):

Since \( g'(x) \neq h(x) \), the function \( g(x) = 5 + e^{2x} + \ln{6} \) is not a primitive function of \( h(x) = 3e^{2x} + \ln{4} \).

Therefore, the answer is:

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