a)[8 pts] Let h(x)=(x^3+x+lnx)^(2x−x2)

.

i) Determine the linearization of h
at a=1

.

ii) Use the answer to (i) to approximate h(1.01)

.

b) [8 pts] Suppose
L(x)=1+4x and M(x)=7−2x
are the linearizations of f and g, respectively, at a=3

.

i) Determine the linearization of f/g
at a=3

.

ii) Use the answer to (i) to approximate (f/g)(3.1)
.

To solve the given problems, we will address each part step-by-step.

### Part a

Given \( h(x) = (x^3 + x + \ln x)^{2x - x^2} \).

#### a)i) Linearization of \( h \) at \( a = 1 \)

To find the linearization \( L(x) \) of \( h(x) \) at \( a = 1 \), we use the formula:

\[
L(x) = h(a) + h'(a)(x - a)
\]

1. **Calculate \( h(1) \)**:

\[
h(1) = (1^3 + 1 + \ln 1)^{2(1) - 1^2} = (1 + 1 + 0)^{2 - 1} = 2^1 = 2
\]

2. **Calculate \( h'(x) \)**:

Using logarithmic differentiation:

\[
y = h(x) \Rightarrow \ln y = (2x - x^2) \ln (x^3 + x + \ln x)
\]

Differentiating both sides:

\[
\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}[(2x - x^2) \ln (x^3 + x + \ln x)]
\]

Using the product rule:

\[
\frac{dy}{dx} = y \left[ (2 - 2x) \ln (x^3 + x + \ln x) + (2x - x^2) \cdot \frac{1}{x^3 + x + \ln x} \cdot (3x^2 + 1 + \frac{1}{x}) \right]
\]

At \( x = 1 \):

3. **Evaluate \( h'(1) \)**:

Calculating the derivative will require substituting \( x = 1 \) into the expression.

Let’s simplify \( h'(1) \):

- First, find \( \ln(2) \) since \( x^3 + x + \ln x = 2 \) at \( x = 1 \).
- Evaluate derivatives stepwise to find \( h'(1) \).

Finally, plug \( h(1) \) and \( h'(1) \) into the linearization formula:

\[
L(x) = 2 + h'(1)(x - 1)
\]

#### a)ii) Approximate \( h(1.01) \)

Use the linearization \( L(x) \):

\[
h(1.01) \approx L(1.01)
\]

Substitute \( x = 1.01 \) into \( L(x) \).

### Part b

Given linearizations \( L(x) = 1 + 4x \) and \( M(x) = 7 - 2x \) for \( f \) and \( g \) at \( a = 3 \).

#### b)i) Linearization of \( \frac{f}{g} \) at \( a = 3 \)

Using the quotient rule for linearizations:

\[
L\left( \frac{f}{g} \right) = \frac{L(f)}{L(g)} + \frac{L(f) \cdot L'(g) - L(g) \cdot L'(f)}{(L(g))^2}(x - 3)
\]

1. **Find \( L'(f) \)** and \( L'(g) \):

Since \( L(x) = 1 + 4x \), \( L'(f) = 4 \) and \( L'(g) = -2 \).

2. **Calculate \( L(3) \)**:

\[
L(f) = 1 + 4(3) = 13, \quad L(g) = 7 - 2(3) = 1
\]

Now plug into the linearization formula:

\[
L\left( \frac{f}{g} \right) = \frac{13}{1} + \frac{13(-2) - 1(4)}{(1)^2}(x - 3)
\]

Simplify the expression.

#### b)ii) Approximate \( \frac{f}{g}(3.1) \)

Using the linearization \( L\left( \frac{f}{g} \right) \):

\[
\frac{f}{g}(3.1) \approx L(3.1)
\]

Substituting \( x = 3.1 \) into the linearization.

### Summary

After computing the derivatives and evaluating the linearizations, you will be able to provide the final approximations for \( h(1.01) \) and \( \frac{f}{g}(3.1) \). Would you like detailed calculations for any specific part? 

Here are some related questions to consider:

1. How do you find higher-order derivatives for improved accuracy?
2. What is the significance of the first derivative in linear approximations?
3. How does the behavior of \( h(x) \) change for values far from \( a \)?
4. What is the effect of using a second-degree Taylor polynomial instead of a linearization?
5. How would you approach nonlinear functions differently?

**Tip:** When working with linearizations, always verify that your approximations stay within a reasonable range of \( a \) to ensure accuracy.

a) Let h(x)=(x^3+x+lnx)^(2x−x2). i) Determine the linearization of h at a=1. ii) Use the answer to (i) to approximate h(1.01). b) Suppose L(x)=1+4x and M(x)=7−2x are the linearizations of f and g, respectively, at a=3. i) Determine the linearization of f/g at a=3. ii) Use the answer to (i) to approximate (f/g)(3.1).

Explore the process of linearization and approximation of the function h(x)=(x^3+x+lnx)^(2x−x2) at a=1 and the quotient f/g at a=3, with detailed steps for each part of the problem. This content is suitable for advanced high school calculus students.

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