To estimate the value of $(3.03)^3 \times (-3.03)^2$ using a linear approximation, we can utilize the concept of a tangent line approximation. Here's the general process:

Step 1: Define the function

Let’s define a function $f(x) = x^3 \times (-x)^2$ (this is equivalent to $f(x) = -x^5$).

We aim to estimate $f(3.03)$, so we'll use linear approximation at a point close to 3.03. The simplest choice for a point is $x = 3$, since 3.03 is very close to 3.

Step 2: Compute the linear approximation

The linear approximation formula is:

$f(x) \approx f(a) + f'(a)(x - a)$

Where:

• $f(a)$ is the value of the function at $a$,
• $f'(a)$ is the derivative of the function evaluated at $a$,
• $x$ is the point at which we want to estimate the function's value.

Here, we will use $a = 3$ and $x = 3.03$.

1. Compute $f(3)$:

$f(3) = -(3)^5 = -243$

2. Compute the derivative $f'(x)$:

$f'(x) = \frac{d}{dx}(-x^5) = -5x^4$ So, $f'(3) = -5(3)^4 = -5 \times 81 = -405$

3. Compute the linear approximation at $x = 3.03$:

$f(3.03) \approx f(3) + f'(3)(3.03 - 3)$ $f(3.03) \approx -243 + (-405)(0.03)$ $f(3.03) \approx -243 - 12.15 = -255.15$

Step 3: Compute the exact value of $f(3.03)$

Now let’s compute the exact value of $(3.03)^3 \times (-3.03)^2$.

$(3.03)^3 = 27.789927$ $(-3.03)^2 = 9.1809$ $f(3.03) = (3.03)^3 \times (-3.03)^2 = 27.789927 \times 9.1809 \approx -255.397$

Step 4: Calculate the percentage error

The percentage error is given by:

$\text{Percentage Error} = \left| \frac{\text{Exact Value} - \text{Approximate Value}}{\text{Exact Value}} \right| \times 100$

Substituting the values: $\text{Percentage Error} = \left| \frac{-255.397 - (-255.15)}{-255.397} \right| \times 100$ $\text{Percentage Error} = \left| \frac{-0.247}{-255.397} \right| \times 100 \approx 0.097$

Final Answer:

The linear approximation to $(3.03)^3 \times (-3.03)^2$ is $-255.15$, and the percentage error is approximately 0.097%.

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Would you like more details on any of the steps, or is there something else you'd like to explore?

Related questions:

1. How can you use linear approximation for other types of functions (e.g., trigonometric functions)?
2. What is the difference between linear approximation and higher-order approximations?
3. How do you compute the Taylor series expansion for a function?
4. What are some practical applications of linear approximations in physics or engineering?
5. Can you estimate the value of $(3.05)^3 \times (-3.05)^2$ using the same method?

Tip:

Linear approximations are very useful for estimating values of complicated expressions when the point of expansion is close to the value you’re interested in. The closer the point of approximation to the value, the more accurate the estimate!