Question

I et \(f\) be a function such that \(f(x+y)=f(x)+f(y)\) for all \(x\) and \(y\) and \(f(x)=\left(2 x^{2}+3 x\right) g(x)\) for all \(x\) where \(g(x)\) is continuous and \(g(0)=3 .\) Then \(f^{\prime}(x)\) is equal to (A) 9 (B) 3 (C) 6 (D) nonc

Short answer

Answer: The value of \(f'(x)\) is 9.

Step-by-step solution

Find the derivative of \(f(x)\) using the product rule

Recall that the derivative of a product of functions is given by the product rule: \((u\cdot v)'=u'\cdot v+u\cdot v'\) Here, let \(u=2x^2+3x\) and \(v=g(x)\). Then, we have \(u'=4x+3\) and \(v'=g'(x)\). Following the product rule: \(f'(x)=(2x^2+3x)'\cdot g(x) + (2x^2+3x)\cdot g'(x)= (4x+3)g(x) + (2x^2+3x)g'(x)\)

Plug in the given information for \(g(x)\)

We are given that \(g(0)=3\). To find the \(f'(x)\) term involving \(g(0)\), we need to plug in \(x=0\): \(f'(0)=(4\cdot0+3)g(0) + (2\cdot0^2+3\cdot0)g'(0)=3g(0)=3\cdot3=9\) So, \(f'(0)=9\).

Find the continuity condition for \(g(x)\)

Since \(g(x)\) is continuous, the limit as \(x\) approaches 0 for the derivative \(f'(x)\) exists and is equal to the derivative at 0: \(\lim_{x\rightarrow0} (4x+3)g(x) = 9\) Since \(g(0)=3\), we have: \(\lim_{x\rightarrow0} (4x+3)g(x) = (4\cdot0+3)g(0)=9\) Thus, both \(f'(0)\) and the limit of the first term in the derivative have a value of 9.

Analyze \(f'(x)\)

Our derivative is as follows: \(f'(x)=(4x+3)g(x) + (2x^2+3x)g'(x)\) We have found that the first term, \((4x+3)g(x)\), has a limit as \(x\) approaches 0 which is 9, and that \(f'(0)=9\). Hence, the second term, \((2x^2+3x)g'(x)\), must be 0 for all \(x\), because it does not affect the overall value of the derivative, \(f'(x)\). As a result, we can conclude that: \(f'(x)=(4x+3)g(x)\) And we know that \(f'(0)=9\). Therefore, the value of \(f'(x)\) is \(\boxed{9\ \text{(A)}}\).