Originally Posted by
islam hello there
in order to compute the tangent line u need
the value of the function f(2g(x)) at x=3
and the value of the derivative of the function
at the same point meaning F'(3)
first u have to find f(2g(x))
just substitute in f(x) 2g(x)
u get
$\displaystyle 16g(x)^4+8g(x)^3+1$
now what we need is actually
$\displaystyle 16g(3)^4+8g(3)^3+1$
to know the value of this expression we need
to know the value of g(3)
since g(x) is the inverse of f(x)
then f(g(3))=3
in other words:
$\displaystyle g(3)^4+g(3)^3+1=3$
$\displaystyle g(3)^4+g(3)^3-2=0$
by guessing we get that the solution is:
g(3)=1
(there is another solution but its hard to get
it actually we need approximation)
now that we have g(3)=1
then F(3)=16*1^4+8*1^3+1=25
NOW we have to know the value of
F'(3)(the derivative)
$\displaystyle F'(3)=4*16*g(3)^3*g'(3)+3*8*g(3)^2*g'(3)$
to now the value of F'(3) we have to
know the value of g'(3) since we already know the value
of g(3)
now since f(g(x)=x
if we derive both sides using the chain rule
we get
$\displaystyle f'(g(x)*g'(x)=1$
now divide both sides by f'(g(x))
$\displaystyle g'(x)=1/f'(g(x)$
$\displaystyle g'(x)=1/f'(g(x)$
$\displaystyle g'(x)=1/(4g(x)^3+3g(x)^2)$
now just subsitute x=3
u get
g'(3)=1/7
can u proceed from here??