Question

A differentiable function f(x) is defined such that, for all values of x in its domain, (i will draw the equation).

(a) what is the domain.
(b) for what values of x is f(x)=3?
(c)show that f'(x)=3^(2)f(x)
(d)solve the differential equation in (c) to find f(x) in terms of x only.

Answer 1

\[f(x)=3+\int_8^{x^3}f(\sqrt[3]{t})dt\] like that ?

Answer 2

yes.

Answer 3

looks to me like the domian is all real numbers

Answer 4

i got that part but rest of the questions are confusing.lol

Answer 5

also i am thinking it should be
\[F(x)=3+\int_8^{x^3}f(\sqrt[3]{t})dt\]

Answer 6

since it is clearly not the same F and f

Answer 7

both say f

Answer 8

\[F(x)=3\implies \int_8^{x^3}f(\sqrt[3]{x})dt=0\]

Answer 9

integral is zero when upper and lower limits of integration are the same, so you have
\[\int_8^{x^3}f(\sqrt[3]{t})dt=\int_8^8f(\sqrt[3]{t})dt\] and so
\[x^3=8\] thefore \(x=2\)

Answer 10

since the derivative of the integral is the integrand, you have
\[f'(x)=f(\sqrt[3]{x^3})\times 3x^2=3x^2f(x)\]

Answer 11

so i am hoping there is a typo above and you meant
\[ f'(x)=3x^2f(x)\]not
\[ f'(x)=3^2f(x)\]

Answer 12

yes it was a typo sorry i didnt catch it.

Answer 13

but it is clear yes?

Answer 14

yes it is thank you.

Answer 15

does that mean that d is f(x)=3ex^3-8

Answer 16

ooh lets see
we didn't finish

Answer 17

if \(f(x)=e^{x^3}\) then
\[f'(x)=3x^2e^{x^3}\] so i don't think you need the additional 3 out front

Answer 18

oh damn we also have to make sure that \(f(2)=3\) don't we

Answer 19

maybe i am confused, but i think we have \(f(x)=e^{x^3}+C\) and since \(f(2)=3\) we should solve
\[3=e^8+C\] for C to get \(C=3-e^8\) but i could be wrong

Answer 20

\[3ex^{3}-8\]thats why i thought it would be