IDNLearn.com provides a seamless experience for finding and sharing answers. Find the information you need quickly and easily with our comprehensive and accurate Q&A platform.
Sagot :
Sure, let's find the linear approximations for the given functions for values of [tex]\( x \)[/tex] near zero using the approximation [tex]\( (1+x)^k \approx 1 + kx \)[/tex].
### Function 1: [tex]\( f(x) = (1-x)^5 \)[/tex]
For this function, we can use the linear approximation [tex]\( (1+x)^k \approx 1 + kx \)[/tex], where [tex]\( k = 5 \)[/tex] and [tex]\( x \)[/tex] is replaced by [tex]\(-x\)[/tex].
So,
[tex]\[ (1-x)^5 \approx 1 + (-5)x = 1 - 5x. \][/tex]
Thus, the approximation for [tex]\( f(x) = (1-x)^5 \)[/tex] is:
[tex]\[ f(x) \approx 1 - 5x. \][/tex]
### Function 2: [tex]\( f(x) = \frac{7}{1-x} \)[/tex]
Here, we can start by approximating the term [tex]\( \frac{1}{1-x} \)[/tex]. Using the approximation [tex]\( \frac{1}{1-x} \approx 1 + x \)[/tex],
Then, multiplying by 7,
[tex]\[ \frac{7}{1-x} \approx 7 (1 + x) = 7 + 7x. \][/tex]
Thus, the approximation for [tex]\( f(x) = \frac{7}{1-x} \)[/tex] is:
[tex]\[ f(x) \approx 7 + 7x. \][/tex]
### Function 3: [tex]\( f(x) = (4+3x)^{1/3} \)[/tex]
For this function, let’s perform a change of variable. Define [tex]\( u = 4 + 3x \)[/tex], we need to approximate [tex]\( u^{1/3} \)[/tex] around [tex]\( u = 4 \)[/tex].
Using the linear approximation [tex]\( (1 + y)^k \approx 1 + ky \)[/tex], and setting [tex]\( y = \frac{3x}{4} \)[/tex], we have:
[tex]\[ (1 + \frac{3x}{4})^{1/3} \approx 1 + \frac{1}{3} \cdot \frac{3x}{4} = 1 + \frac{x}{4}. \][/tex]
Since the approximation should be made around 4, we need to multiply [tex]\( 4^{1/3} \)[/tex] by the linear approximation. Therefore,
[tex]\[ (4 + 3x)^{1/3} \approx 4^{1/3} (1 + \frac{x}{4}) = 4^{1/3} + \frac{4^{1/3} x}{4}. \][/tex]
Thus, the approximation for [tex]\( f(x) = (4+3x)^{1/3} \)[/tex] is approximately:
[tex]\[ f(x) \approx 4^{1/3} (1 + \frac{x}{4}).\][/tex]
So, putting together the solutions:
1. [tex]\( f(x) = (1-x)^5 \approx 1 - 5x \)[/tex].
2. [tex]\( f(x) = \frac{7}{1-x} \approx 7 + 7x \)[/tex].
3. [tex]\( f(x) = (4+3x)^{1/3} \approx 4^{1/3} (1 + \frac{x}{4}).\)[/tex]
### Function 1: [tex]\( f(x) = (1-x)^5 \)[/tex]
For this function, we can use the linear approximation [tex]\( (1+x)^k \approx 1 + kx \)[/tex], where [tex]\( k = 5 \)[/tex] and [tex]\( x \)[/tex] is replaced by [tex]\(-x\)[/tex].
So,
[tex]\[ (1-x)^5 \approx 1 + (-5)x = 1 - 5x. \][/tex]
Thus, the approximation for [tex]\( f(x) = (1-x)^5 \)[/tex] is:
[tex]\[ f(x) \approx 1 - 5x. \][/tex]
### Function 2: [tex]\( f(x) = \frac{7}{1-x} \)[/tex]
Here, we can start by approximating the term [tex]\( \frac{1}{1-x} \)[/tex]. Using the approximation [tex]\( \frac{1}{1-x} \approx 1 + x \)[/tex],
Then, multiplying by 7,
[tex]\[ \frac{7}{1-x} \approx 7 (1 + x) = 7 + 7x. \][/tex]
Thus, the approximation for [tex]\( f(x) = \frac{7}{1-x} \)[/tex] is:
[tex]\[ f(x) \approx 7 + 7x. \][/tex]
### Function 3: [tex]\( f(x) = (4+3x)^{1/3} \)[/tex]
For this function, let’s perform a change of variable. Define [tex]\( u = 4 + 3x \)[/tex], we need to approximate [tex]\( u^{1/3} \)[/tex] around [tex]\( u = 4 \)[/tex].
Using the linear approximation [tex]\( (1 + y)^k \approx 1 + ky \)[/tex], and setting [tex]\( y = \frac{3x}{4} \)[/tex], we have:
[tex]\[ (1 + \frac{3x}{4})^{1/3} \approx 1 + \frac{1}{3} \cdot \frac{3x}{4} = 1 + \frac{x}{4}. \][/tex]
Since the approximation should be made around 4, we need to multiply [tex]\( 4^{1/3} \)[/tex] by the linear approximation. Therefore,
[tex]\[ (4 + 3x)^{1/3} \approx 4^{1/3} (1 + \frac{x}{4}) = 4^{1/3} + \frac{4^{1/3} x}{4}. \][/tex]
Thus, the approximation for [tex]\( f(x) = (4+3x)^{1/3} \)[/tex] is approximately:
[tex]\[ f(x) \approx 4^{1/3} (1 + \frac{x}{4}).\][/tex]
So, putting together the solutions:
1. [tex]\( f(x) = (1-x)^5 \approx 1 - 5x \)[/tex].
2. [tex]\( f(x) = \frac{7}{1-x} \approx 7 + 7x \)[/tex].
3. [tex]\( f(x) = (4+3x)^{1/3} \approx 4^{1/3} (1 + \frac{x}{4}).\)[/tex]
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. IDNLearn.com is committed to providing the best answers. Thank you for visiting, and see you next time for more solutions.