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a) Find the linear approx for the following centered at a b) find the quadratic approx polynomial...

Question:

a) find the linear approx for the following centered at a

b) find the quadratic approx polynomial for the following centered at a

c) use polynomials from parts (a) and (b) to approx the given quantity.

{eq}f(x) = 8 x^{(\frac{3}{4})}, a = 1: \ approx \ 8 \times 1.1^ \frac{3}{2}. {/eq}

Linear and Quadratic Approximations:

The linear approximation of a function {eq}f(x) {/eq} at the point {eq}x=x_0 {/eq} is found as

{eq}L(x) = f(x_0) + f'(x_0)(x-x_0). {/eq}

The quadratic approximation at the same point is

{eq}Q(x) = f(x_0) + f'(x_0)(x-x_0) + \frac{1}{2}f''(x_0)(x-x_0)^2. {/eq}

Answer and Explanation:

We are given the function

{eq}\displaystyle f(x) = 8 x^{\frac{3}{4}}. {/eq}

a) The linear approximation of the function centered at {eq}x =1 {/eq} is calculated as follows

{eq}f(1) = 8 \\ \displaystyle f'(x) = 6 x^{-\frac{1}{4}} \Rightarrow f'(1)=6 \\ \displaystyle L(x) = f(1) + f'(1)(x-1)= 8 + 6(x-1) = 6x+2. {/eq}

b) The quadratic approximation of the function centered at {eq}x =1 {/eq} is calculated as follows

{eq}\displaystyle f(1) = 8 \\ \displaystyle f'(x) = 6 x^{-\frac{1}{4}} \Rightarrow f'(1)=6 \\ \displaystyle f''(x) = -\frac{3}{2} x^{-\frac{5}{4}} \Rightarrow f''(1)= -\frac{3}{2} \\ \displaystyle Q(x) = f(1) + f'(1)(x-1) + \frac{1}{2}f''(1)(x-1)^2 = 8 + 6(x-1) -\frac{3}{4}(x-1)^2 \\ \displaystyle Q(x) = -\frac{3}{4}x^2+\frac{15}{2}x+\frac{5}{4} {/eq}

c) Using polynomials from parts a and b, we approximate the given quantity:

{eq}\displaystyle 8 \times 1.1^ \frac{3}{2} = 8 \times 1.21^ \frac{3}{4} \approx L(1.21) = 9.26 \\ \displaystyle 8 \times 1.1^ \frac{3}{2} = 8 \times 1.21^ \frac{3}{4} \approx Q(1.21) = 9.23. {/eq}


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Linearization of Functions

from Math 104: Calculus

Chapter 10 / Lesson 1
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