Question

step 1
remember that it is not true that if f and g are antiderivatives of f and g, respectively, then f·g is an antiderivative of f·g. however, we can expand the product and use fractional exponents to rewrite ( f(x)=sqrt{x}(9 + 5x) ) as
( 9x^{1/2}+5x^{3/2} ).

step 2
now, the most general antiderivative of ( f(x)=9x^{1/2}+5x^{3/2} ) is
( f(x)=6x^{(\frac{3}{2})}+2x^{(\frac{5}{2})}+c ).

step 3
we will use the condition that ( f(1)=12 ) to find c. we have
( f(1)=6(quad)^{3/2}+2(quad)^{5/2}+c ).
therefore,
( 12=quad + c )
( quad = c ).

Explanation:

Step1: Substitute \(x = 1\) into \(f(x)\)

We know \(f(x)=6x^{\frac{3}{2}}+2x^{\frac{5}{2}}+C\). Substituting \(x = 1\) into \(f(x)\), we get \(f(1)=6(1)^{\frac{3}{2}}+2(1)^{\frac{5}{2}}+C\).

Step2: Simplify the expression

Since \(1^n=1\) for any real number \(n\), \(6(1)^{\frac{3}{2}}+2(1)^{\frac{5}{2}}=6\times1 + 2\times1=6 + 2=8\). So, from \(12=8 + C\).

Step3: Solve for \(C\)

Subtracting \(8\) from both sides of the equation \(12=8 + C\), we get \(C=12-8 = 4\).

Answer:

The first blank is \(1\), the second blank is \(1\), the third blank is \(8\), and the fourth blank is \(4\).