By the quotient and product rules,
[tex]\left(\dfrac fg\right)'(0) = \dfrac{g(0) f'(0) - f(0) g'(0)}{g(0)^2} = 1[/tex]
[tex](f\times g)'(0) = f(0) g'(0) + f'(0) g(0) = 21[/tex]
Given that [tex]f'(0)=5[/tex] and [tex]g'(0)=3[/tex], we have the system of equations
[tex]\dfrac{5g(0) - 3f(0)}{g(0)^2} = 1 \implies 5g(0) - 3f(0) = g(0)^2[/tex]
[tex]3f(0) + 5g(0) = 21[/tex]
Eliminating [tex]f(0)[/tex] gives
[tex]\bigg(5g(0) - 3f(0)\bigg) + \bigg(3f(0) + 5g(0)\bigg) = g(0)^2 + 21[/tex]
[tex]10g(0) = g(0)^2 + 21[/tex]
[tex]g(0)^2 - 10g(0) + 21 = 0[/tex]
[tex]\bigg(g(0) - 7\bigg) \bigg(g(0) - 3\bigg) = 0[/tex]
[tex]\implies \boxed{g(0) = 7 \text{ or } g(0) = 3}[/tex]
Solve for [tex]f(0)[/tex].
[tex]3f(0) + 5g(0) = 21[/tex]
[tex]3f(0) + 35 = 21 \text{ or } 3f(0) + 15 = 21[/tex]
[tex]3f(0) = -14 \text{ or } 3f(0) = 6[/tex]
[tex]\implies \boxed{f(0) = -\dfrac{14}3 \text{ or } 3f(0) = 2}[/tex]
