Question

Matching functions with polynomials Match functions a–f with Taylor polynomials A–F (all centered at 0). Give reasons for your choices.

a. √(1 + 2x)

A. p₂(x)= 1 + 2x + 2x²
B. p₂(x) = 1 − 6x + 24x²
C. p₂(x) = 1 + x − x²/2
D. p₂(x) = 1 − 2x + 4x²
E. p₂(x) = 1 − x + (3/2)x²
F. p₂(x) = 1 − 2x + 2x²

Accepted Answer

Identify the function to approximate: $$f(x) = \sqrt{1 + 2x} = (1 + 2x$$)^{1/2}$$, and note that the Taylor polynomials are centered at 0 (Maclaurin polynomials). Recall the general formula for the Taylor polynomial of degree 2 centered at 0:

p_2$(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2$. Calculate the derivatives of f(x$$)$$:

- First, f(x) = (1$$ + $$2x)^{1/2}.
- $$Use the chain rule for the first derivative:

$$f'(x) = \frac{1}{2}(1 + 2x$$)^{-1/2}$$ \cdot 2 = (1 + 2x$$)^{-1/2}$$.

- For the second derivative, differentiate f$$'(x)$$:

f$$''(x) = -\frac{1}{2}(1 + $$2x)^{-3/2}$$ \cdot 2 = - (1 + $$2x)^{-3/2}. $$Evaluate the function and its derivatives at $$x=0$$:

- $$f(0) = (1 + 0$$)^{1/2}$$ = 1$$,
- $$f'(0) = (1 + 0$$)^{-1/2}$$ = 1$$,
- $$f''(0) = - (1 + 0$$)^{-3/2}$$ = -1. $$Substitute these values into the Taylor polynomial formula:

$$p_2(x) = 1 + 1$$ \cdot x + \frac{-1}{2} $$x^2 = 1 + x$$ - \frac{1}{2} $$x^2.

$$Compare this polynomial with the given options to find the matching polynomial.

Verified video answer for a similar problem:

Key Concepts

Taylor Polynomials

Derivatives and Their Role in Taylor Series

Matching Functions to Polynomials via Coefficients