3.2 Exponential Function best fitting f(x) = e^x at x = 0

[tglass]

#43. Find the quadratic polynomial g(x) = ax^2+bx+c which best fits the function f(x)=e^x at x=0 in the sense that:
g(0)=f(0) and g'(0)=f'(0) and g"(0)=f"(0).

so, because e^0=1, do I set g(x)=1? I don't understand even where to begin on this one. f'(x) would be 1 (because f'(x)=e^x still). So all I know is the slope and the points (0,1). How do I develop a quadratic out of that? (y-1)=1 (x-0); y=x+1...that's the equation for the line tanget to e^x, no?

As always, thanks in advance,
Tim

 

[Deleted member 4993]

Re: 3.2 Exponential Function

#43.
Find the quadratic polynomial g(x) = ax^2+bx+c which best fits the function f(x)=e^x at x=0 in the sense that:
g(0)=f(0) and g'(0)=f'(0) and g"(0)=f"(0).

so, because e^0=1, do I set g(x)=1? <<< No you set g(0) = 1 -> that should give you value of 'c'.

I don't understand even where to begin on this one.

What is the expression for g'(x)?

What is the value of g'(0) - in terms of 'a' , 'b' & 'c'?

Now equate that to f'(0) - and see if you can solve for one or more constants (such as 'a', 'b' or 'c')


f'(x) would be 1 (because f'(x)=e^x still). So all I know is the slope and the points (0,1). How do I develop a quadratic out of that? (y-1)=1 (x-0); y=x+1...that's the equation for the line tanget to e^x, no?

As always, thanks in advance,
Tim

 

[BigGlenntheHeavy]

Re: 3.2 Exponential Function

f(x) = e^x, g(x) = ax^2+bx+c
f(0)=1 g(0) = c, hence =1
f' (0) = 1 g' (0) =b, hence b =1
f''(0) =1 g''(0 ) =2a, a = 1/2

Ergo g(0) =(x^2/2)+x+1 is the closest you'll get to f(x) with the above restrictions. x=0.

f(0) = 1 =g(0),c=1

f' (0) =1 =g' (0), b=1

f''(0) = 1 =g''(0), 2a =1, a=1/2