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]
