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Let's go through each part of this problem step-by-step.

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### Part (a): Let \( h(x) = (x^3 + x + \ln x)^{2x - x^2} \)

**(i) Determine the linearization of \( h \) at \( a = 1 \).**

To find the linearization \( L(x) \) of \( h(x) \) at \( x = 1 \), we use the formula:
\[
L(x) = h(1) + h'(1)(x - 1)
\]
So, we need to find \( h(1) \) and \( h'(1) \).

1. **Calculate \( h(1) \):**
   Substitute \( x = 1 \) into \( h(x) \):
   \[
   h(1) = (1^3 + 1 + \ln 1)^{2 \cdot 1 - 1^2} = (1 + 1 + 0)^{2 - 1} = 2^1 = 2
   \]

2. **Calculate \( h'(x) \):**
   Since \( h(x) = (x^3 + x + \ln x)^{2x - x^2} \), we use the logarithmic differentiation technique to differentiate.

   Let \( u(x) = x^3 + x + \ln x \) and \( v(x) = 2x - x^2 \). Then,
   \[
   h(x) = u(x)^{v(x)}
   \]
   Taking the natural logarithm,
   \[
   \ln h(x) = v(x) \ln u(x)
   \]
   Differentiating both sides with respect to \( x \):
   \[
   \frac{h'(x)}{h(x)} = v'(x) \ln u(x) + v(x) \frac{u'(x)}{u(x)}
   \]
   Solving this for \( h'(x) \) and evaluating at \( x = 1 \) will yield \( h'(1) \).

**(ii) Approximate \( h(1.01) \) using the linearization.**

Using \( L(x) \) from part (i), substitute \( x = 1.01 \) into \( L(x) \) to approximate \( h(1.01) \).

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### Part (b): Suppose \( L(x) = 1 + 4x \) and \( M(x) = 7 - 2x \) are the linearizations of \( f \) and \( g \) at \( a = 3 \).

**(i) Determine the linearization of \( \frac{f}{g} \) at \( a = 3 \).**

The linearization of \( \frac{f}{g} \) at \( x = 3 \) can be found using the formula for the derivative of a quotient:
\[
\left( \frac{f}{g} \right)' = \frac{f'g - fg'}{g^2}
\]
Using the values of \( f(3) \), \( f'(3) \), \( g(3) \), and \( g'(3) \) derived from \( L(x) \) and \( M(x) \), compute \( \frac{f}{g} \) at \( x = 3 \).

**(ii) Use the answer to (i) to approximate \( \frac{f}{g}(3.1) \).**

With the linearization found in (i), substitute \( x = 3.1 \) to approximate \( \frac{f}{g}(3.1) \).

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Would you like detailed calculations for each derivative and linear approximation? 

### Related Questions

1. How would you calculate \( h'(x) \) using implicit differentiation?
2. Can you linearize a function with two variables, such as \( h(x, y) \)?
3. What are the steps for finding higher-order approximations beyond linearization?
4. How does linearization relate to Taylor series expansion?
5. What are some applications of linear approximations in physics or engineering?

### Tip
When working with exponential functions, logarithmic differentiation can simplify the process by allowing you to work with products instead of powers.

Let h(x) = (x^3 + x + ln x)^{2x - x^2}. Determine the linearization of h at a = 1, and use it to approximate h(1.01). Suppose L(x) = 1 + 4x and M(x) = 7 - 2x are the linearizations of f and g, respectively, at a = 3. Determine the linearization of f/g at a = 3 and use it to approximate (f/g)(3.1).

This problem involves determining the linearization of h(x) = (x^3 + x + ln x)^{2x - x^2} at a = 1 and using it to approximate h(1.01). Additionally, we explore the linearizations of f and g to find the linearization of f/g at a = 3 and approximate (f/g)(3.1). Key concepts include differentiation techniques and approximation methods.

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