# What is a,b,c if f(x)=e^x(ax^2+bx+c), f(0)=0,f'(0)=1,f''(0)=4?

### 1 Answer | Add Yours

You need to evaluate the function, the first and the second derivatives at `x = 0` , such that:

`f(0) = e^0(a*0^2 + b*0 + c) => f(0) = c`

`f(0) = 0 => c = 0`

You need to evaluate the first derivative of the function using the product rule, such that:

`f'(x) = (e^x)'(ax^2 + bx + c) + (e^x)(ax^2 + bx + c)'`

`f'(x) = (e^x)(ax^2 + bx + c) + (e^x)(2ax + b)`

Factoring out `e^x` yields:

`f'(x) = e^x(ax^2 + bx + c + 2ax + b)`

You need to evaluate `f'(x)` at `x = 0` , suxh that:

`f'(0) = e^0(a*0^2 + b*0 + 0 + 2a*0 + b)`

`f'(0) = 1*b => b = 1`

You need to evaluate the second derivative of the function using the product rule, such that:

`f''(x) = (e^x)'(ax^2 + bx + c + 2ax + b) + (e^x)(ax^2 + bx + c + 2ax + b)'`

`f''(x) = (e^x)(ax^2 + bx + c + 2ax + b) + (e^x)(2ax + b + 2a)`

`f''(x) = (e^x)(ax^2 + bx + c + 2ax + b + 2ax + b + 2a)`

`f''(x) = (e^x)(ax^2 + bx + c + 4ax + 2b + 2a)`

You need to evaluate `f''(x)` at `x = 0` , suxh that:

`f''(0) = (e^0)(2 + 2a) => 4 = 2 + 2a => 2(1 + a) = 4 => 1 + a = 2 => a = 1`

**Hence, evaluating the coefficients a,b,c, under the given conditions, yields `a = 1, b = 1, c = 0` .**