Question

Find f(0) and g(0), if (f/g)'(0)=1, (f g)'(0)=21, f'(0)=5,g'(0)=3

Answer 1

By the quotient and product rules,

`\left(\frac fg\right)'(0) = (g(0) f'(0) - f(0) g'(0))/(g(0)^2) = 1`

`(f* g)'(0) = f(0) g'(0) + f'(0) g(0) = 21`

Given that
`f'(0)=5` and
`g'(0)=3`, we have the system of equations

`(5g(0) - 3f(0))/(g(0)^2) = 1 \implies 5g(0) - 3f(0) = g(0)^2`

`3f(0) + 5g(0) = 21`

Eliminating
`f(0)` gives

`\bigg(5g(0) - 3f(0)\bigg) + \bigg(3f(0) + 5g(0)\bigg) = g(0)^2 + 21`

`10g(0) = g(0)^2 + 21`

`g(0)^2 - 10g(0) + 21 = 0`

`\bigg(g(0) - 7\bigg) \bigg(g(0) - 3\bigg) = 0`

`\implies \boxed{g(0) = 7 \text{ or } g(0) = 3}`

Solve for
`f(0)`.

`3f(0) + 5g(0) = 21`

`3f(0) + 35 = 21 \text{ or } 3f(0) + 15 = 21`

`3f(0) = -14 \text{ or } 3f(0) = 6`

`\implies \boxed{f(0) = -\frac{14}3 \text{ or } 3f(0) = 2}`