Question

if ( f(x)=e^{2x}(x^{3}+1) ), then ( f(2)= )
a ( 6e^{4} )
b ( 21e^{4} )
c ( 24e^{4} )
d ( 30e^{4} )

Explanation:

Step1: Apply the product rule

The product rule states that if \(y = u(x)v(x)\), then \(y^\prime=u^\prime(x)v(x)+u(x)v^\prime(x)\). Let \(u(x)=e^{2x}\) and \(v(x)=x^{3}+1\).

First, find \(u^\prime(x)\):
Using the chain - rule, if \(y = e^{ax}\), then \(y^\prime=ae^{ax}\). For \(u(x)=e^{2x}\), \(u^\prime(x)=2e^{2x}\).

Next, find \(v^\prime(x)\):
Using the power rule \((x^{n})^\prime = nx^{n - 1}\), for \(v(x)=x^{3}+1\), \(v^\prime(x)=3x^{2}\).

So, \(f^\prime(x)=u^\prime(x)v(x)+u(x)v^\prime(x)=2e^{2x}(x^{3}+1)+e^{2x}\times3x^{2}\).

Step2: Simplify \(f^\prime(x)\)

Factor out \(e^{2x}\):
\(f^\prime(x)=e^{2x}(2(x^{3}+1)+3x^{2})\).
Expand the expression inside the parentheses:
\(2(x^{3}+1)+3x^{2}=2x^{3}+2 + 3x^{2}\).

Step3: Substitute \(x = 2\)

When \(x = 2\), we have:
\(f^\prime(2)=e^{2\times2}(2\times2^{3}+2+3\times2^{2})\).
Calculate the value inside the parentheses:
\(2\times2^{3}+2+3\times2^{2}=2\times8 + 2+3\times4=16 + 2+12=30\).
\(e^{2\times2}=e^{4}\).
So, \(f^\prime(2)=e^{4}(2\times8 + 2+3\times4)=30e^{4}\).

Answer:

D. \(30e^{4}\)