Question

step 1
remember that it is not true that if ( f ) and ( g ) are antiderivatives of ( f ) and ( g ), respectively, then ( f cdot g ) is an antiderivative of ( f cdot g ). however, we can expand the product and use fractional exponents to rewrite ( f^{prime}(x)=sqrt{x}(9 + 5x) ) as
9 x^{1 / 2}+5 x^{3 / 2}
step 2
now, the most general antiderivative of ( f^{prime}(x)=9 x^{1 / 2}+5 x^{3 / 2} ) is
f(x)=6 x^{left(\frac{3}{2}
ight)}+2 x^{left(\frac{5}{2}
ight)}+c
step 3
we will use the condition that ( f(1)=12 ) to find ( c ). we have
f(1)=6(1)^{3 / 2}+2(1)^{5 / 2}+c
therefore,
12=8 + c 4=c
step 4
putting this together we have that
f(x)=

Explanation:

Step1: Expand the function

We start with \(f^{\prime}(x)=\sqrt{x}(9 + 5x)\). Using the rule \(\sqrt{x}=x^{\frac{1}{2}}\) and the distributive property \(a(b + c)=ab+ac\), we get \(f^{\prime}(x)=9x^{\frac{1}{2}}+5x^{\frac{3}{2}}\).

Step2: Find the antiderivative

The antiderivative of \(x^{n}\) is \(\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)).
For the term \(9x^{\frac{1}{2}}\), using the formula \(\int ax^{n}dx=a\frac{x^{n + 1}}{n + 1}+C\) (\(a = 9\), \(n=\frac{1}{2}\)), we have \(\int9x^{\frac{1}{2}}dx=9\times\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}=9\times\frac{x^{\frac{3}{2}}}{\frac{3}{2}}=6x^{\frac{3}{2}}\).
For the term \(5x^{\frac{3}{2}}\), with \(a = 5\), \(n=\frac{3}{2}\), \(\int5x^{\frac{3}{2}}dx=5\times\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}=5\times\frac{x^{\frac{5}{2}}}{\frac{5}{2}}=2x^{\frac{5}{2}}\).
So, \(f(x)=6x^{\frac{3}{2}}+2x^{\frac{5}{2}}+C\).

Step3: Use the initial - condition

Given \(f(1) = 12\), substitute \(x = 1\) into \(f(x)=6x^{\frac{3}{2}}+2x^{\frac{5}{2}}+C\).
We get \(f(1)=6\times(1)^{\frac{3}{2}}+2\times(1)^{\frac{5}{2}}+C\). Since \(1^{m}=1\) for any real \(m\), then \(12=6 + 2+C\).
Simplifying the right - hand side gives \(12=8 + C\). Solving for \(C\), we subtract 8 from both sides: \(C=12 - 8=4\).

Answer:

\(f(x)=2x^{\frac{5}{2}}+6x^{\frac{3}{2}}+4\)