In this question, we will estimate the value of  (9/10)1/3
 using a linearization of  f(x)=(1+4x)1/3
 .

 

a) Find  f′(0)=
    
4/3
 

 

b) Find the linearization  L(x)
  of  f(x)
  at the point  x=0
 . 

L(x)=
  
1+(4/3)*x
Preview Change entry mode 

FORMATTING: Your answer must be a function of  x
 .

 

c) Now work out for what value of  x
  we have  f(x)=(9/10)1/3
 .

Answer:  x=
    
−0.025
 

 

d) Since your answer in (c) is close to  0
 , we may use our linearization in (b) to estimate  (9/10)1/3
 .

Answer =    
0.9667
  

You may verify with your calculator that this answer is close to the true valueNumeric entries can be read by the system in the following styles:

Optional thousands separators can be used, so that 1,234 and 1234 will both be accepted.
Dollar signs will not be accepted in your answers.
You can use scientific notation in your answer, for example 2.37E-3 in place of 0.00237.
You can enter arithmetic expressions for your answer. For example if you enter 2.35+0.11 the system will grade it as though you had entered 2.46.
Negative numbers are denoted with a minus sign. For example, -27.5.

Let's go through each part of this problem to confirm the solution.

### Part (a): Find \( f'(0) \)

Given:
\[
f(x) = (1 + 4x)^{\frac{1}{3}}
\]

To find \( f'(x) \), we'll use the chain rule.

1. **Differentiate** \( f(x) = (1 + 4x)^{\frac{1}{3}} \):
   \[
   f'(x) = \frac{1}{3}(1 + 4x)^{-\frac{2}{3}} \cdot 4
   \]
   Simplifying, we get:
   \[
   f'(x) = \frac{4}{3} (1 + 4x)^{-\frac{2}{3}}
   \]

2. **Evaluate at \( x = 0 \)**:
   \[
   f'(0) = \frac{4}{3} (1 + 4 \cdot 0)^{-\frac{2}{3}} = \frac{4}{3}
   \]

So, \( f'(0) = \frac{4}{3} \).

---

### Part (b): Find the linearization \( L(x) \) of \( f(x) \) at \( x = 0 \)

The linearization of \( f(x) \) at \( x = 0 \) is given by:
\[
L(x) = f(0) + f'(0) \cdot (x - 0)
\]

1. **Calculate \( f(0) \)**:
   \[
   f(0) = (1 + 4 \cdot 0)^{\frac{1}{3}} = 1
   \]

2. **Substitute values into \( L(x) \)**:
   \[
   L(x) = 1 + \frac{4}{3} \cdot x
   \]

So, \( L(x) = 1 + \frac{4}{3}x \).

---

### Part (c): Find \( x \) such that \( f(x) = \left(\frac{9}{10}\right)^{\frac{1}{3}} \)

We need to find \( x \) such that:
\[
f(x) = (1 + 4x)^{\frac{1}{3}} = \left(\frac{9}{10}\right)^{\frac{1}{3}}
\]

1. **Rewrite in terms of powers**:
   \[
   1 + 4x = \frac{9}{10}
   \]

2. **Solve for \( x \)**:
   \[
   4x = \frac{9}{10} - 1 = -\frac{1}{10}
   \]
   \[
   x = -\frac{1}{40} = -0.025
   \]

So, \( x = -0.025 \).

---

### Part (d): Use linearization to estimate \( \left(\frac{9}{10}\right)^{\frac{1}{3}} \)

Using the linear approximation \( L(x) = 1 + \frac{4}{3} x \) with \( x = -0.025 \):
\[
L(-0.025) = 1 + \frac{4}{3} \cdot (-0.025)
\]
\[
= 1 - \frac{4}{3} \cdot 0.025
\]
\[
= 1 - \frac{0.1}{3}
\]
\[
= 1 - 0.0333 \approx 0.9667
\]

Therefore, the estimated value of \( \left(\frac{9}{10}\right)^{\frac{1}{3}} \) using the linearization is approximately \( 0.9667 \). 

Let me know if you need further details, or if you'd like to check the steps with different values. Here are some related questions:

1. How is the chain rule used to find the derivative in functions involving exponents?
2. Why is linearization useful for approximations near a certain point?
3. What other methods can approximate \( \left(\frac{9}{10}\right)^{\frac{1}{3}} \)?
4. How does the accuracy of the linear approximation vary with distance from \( x = 0 \)?
5. Can we improve this approximation with higher-order terms?

**Tip:** Linearization is especially useful when evaluating functions at points near the point of expansion, as the function behaves similarly to the tangent line at that location.

In this question, we will estimate the value of (9/10)^(1/3) using a linearization of f(x)=(1+4x)^(1/3). a) Find f′(0)= 4/3. b) Find the linearization L(x) of f(x) at the point x=0. L(x)= 1+(4/3)*x. c) Find x such that f(x)=(9/10)^(1/3). d) Use the linearization to estimate (9/10)^(1/3).

This problem demonstrates how to use linearization to estimate the value of (9/10)^(1/3) by finding the derivative and linear approximation of f(x)=(1+4x)^(1/3) at x=0. It covers calculus concepts including linearization, chain rule, and function derivatives, suitable for students in Grades 11-12.

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