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

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 $h(1.01)$.

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 $\frac{f}{g}(3.1)$.

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.