Find the quadratic polynomial g(x)-ax^{2} + bx + c text{which best fits the function} f(x)=e^{x} text{at} x=0, text{in the sense that} g(0)=f(0), text{and} g'(0)=f'(0), text{and} g''(0)=f''(0). Using a computer or calculator, sketch graphs of f and g on the same axes. What do you notice?

Step 1 The functions f(x) and g(x) are: ${\displaystyle f\left(x\right)=e^x}$
${\displaystyle g\left(x\right)=a{x}^2+bx+c}$
${\displaystyle f^{\prime }\left(x\right)=\frac{d\left(f\left(x\right)\right)}{dx}=\frac{d\left(e^x\right)}{dx}}$

Using formula for derivative of exponential function: ${\displaystyle f^{\prime }\left(x\right)=e^x}$
${\displaystyle f{}^{″}\left(x\right)=\frac{d(f,\left(x\right))}{dx}=\frac{d\left(e^x\right)}{dx}}$

Using formula for derivative of exponential function: ${\displaystyle f{}^{″}\left(x\right)=e^x}$
${\displaystyle g^{\prime }\left(x\right)=\frac{d\left(g\left(x\right)\right)}{dx}=\frac{d(a{x}^2+bx+c)}{dx}}$

Using Theorem 3.2: ${\displaystyle g^{\prime }\left(x\right)=\frac{d\left(a{x}^2\right)}{dx}+\frac{d\left(bx\right)}{dx}+\frac{d\left(c\right)}{dx}}$

Using Theorem 3.1: ${\displaystyle g^{\prime }\left(x\right)=a\times \frac{d\left(x^2\right)}{dx}+b\times \frac{d\left(x\right)}{dx}+0}$

Step 2 Using Power Law: ${\displaystyle g^{\prime }\left(x\right)=a\times 2\times x^{2-1}+b\times 1+0}$
${\displaystyle g^{\prime }\left(x\right)=2ax+b}$
${\displaystyle g{}^{″}\left(x\right)=\frac{d\left(g^{\prime }\left(x\right)\right)}{dx}=\frac{d(2ax+b)}{dx}}$

Using Theorem 3.2: ${\displaystyle g{}^{″}\left(x\right)=\frac{d\left(2ax\right)}{dx}+\frac{d\left(b\right)}{dx}}$

Using Theorem 3.1: <