Certainly! Let's proceed step by step to solve the given problem correctly.
### Part 1(a): Linearization [tex]$L(x)$[/tex] at [tex]$x=3$[/tex]
Given the function [tex]\( f(x) = x \ln(x) - 2 \)[/tex]:
1. Evaluate the function at [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = 3 \ln(3) - 2 \][/tex]
After substituting and evaluating, we find:
[tex]\[ f(3) \approx 1.296 \][/tex]
2. Find the derivative of [tex]\( f(x) \)[/tex]:
The derivative of [tex]\( f(x) \)[/tex] is:
[tex]\[ f'(x) = \ln(x) + 1 \][/tex]
3. Evaluate the derivative at [tex]\( x = 3 \)[/tex]:
[tex]\[ f'(3) = \ln(3) + 1 \][/tex]
After substituting and evaluating, we find:
[tex]\[ f'(3) \approx 2.099 \][/tex]
4. Form the linearization [tex]\( L(x) \)[/tex]:
The formula for the linearization of a function [tex]\( f \)[/tex] at [tex]\( x = a \)[/tex] is given by:
[tex]\[ L(x) = f(a) + f'(a) (x - a) \][/tex]
Plugging in the values we found:
[tex]\[ L(x) = 1.296 + 2.099 (x - 3) \][/tex]
5. Simplify the expression:
Simplifying, we get:
[tex]\[ L(x) \approx 2.099x - 6.001 \][/tex]
### Part 1(b): Finding the [tex]\( x \)[/tex]-intercept of [tex]\( L(x) \)[/tex]
To find the [tex]\( x \)[/tex]-intercept of the linearization [tex]\( L(x) \)[/tex], we need to solve [tex]\( L(x) = 0 \)[/tex].
1. Set the linear equation to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 2.099x - 6.001 = 0 \][/tex]
2. Solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{6.001}{2.099} \][/tex]
[tex]\[ x \approx 2.383 \][/tex]
### Summary of Answers
Part 1(a): The linearization [tex]\( L(x) \)[/tex] of the function [tex]\( f(x) = x \ln(x) - 2 \)[/tex] at [tex]\( x = 3 \)[/tex] is:
[tex]\[ L(x) \approx 2.099x - 6.001 \][/tex]
Part 1(b): The [tex]\( x \)[/tex]-intercept of [tex]\( L(x) \)[/tex] is:
[tex]\[ x \approx 2.383 \][/tex]
We have rounded the coefficients to three decimal places as requested. This [tex]\( x \)[/tex]-intercept is an estimate of the [tex]\( x \)[/tex]-intercept for the function [tex]\( f(x) \)[/tex].
