write out the fourth degree taylor polynomial for g(x)=f(x^2+3) about x=0

Question

write out the fourth-degree taylor polynomial for g(x)=f(x^2+3) about x=0

Answer 1

Hey Amelia,

This question may need some clarification, but here's my best shot with what's here.

Remember a Taylor polynomial is of the form:

So in our case, the 4th degree polynomial centered at 0 (meaning we use a=0) would look like:

g(0) + g'(0)x + (g''(0)/2!)x² + (g'''(0)/3!)x³ + (g⁽⁴⁾(0)/4!)x⁴

What makes this tricky, is that they've defined g in terms of an unknown function f.

But, using the chain rule, if g(x) = f(x²+3), then g'(x) = (f'(x²+3))(2x).

You could find the higher derivatives in terms of f by using chain rule and the product rule. Then try plugging in above.

Hope that helps!

Answer 2

my guess is x²+3 represented as taylor about 0 is x²+3

f(0)=3

f'(0)=0

f''(0)=2

f''''(0)=0

f(x)=3+(0/1)x+2x²/2+0(1/3!)

=x²+3