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We are given the function \( f(x) = \sqrt{100x^2 + 25} \), and are tasked with finding the equation of the tangent line at \( x = 1 \), using it to approximate \( f(1.1) \), and then finding the error between the actual function value and the linear approximation.

### Step 1: Find the derivative of \( f(x) \)
To find the equation of the tangent line, we need to compute the derivative \( f'(x) \).

\[
f(x) = \sqrt{100x^2 + 25}
\]

Using the chain rule for the derivative of a square root function:

\[
f'(x) = \frac{1}{2\sqrt{100x^2 + 25}} \times 200x
\]

Simplifying:

\[
f'(x) = \frac{100x}{\sqrt{100x^2 + 25}}
\]

### Step 2: Evaluate \( f'(x) \) at \( x = 1 \)
Now we substitute \( x = 1 \) into the derivative to find the slope of the tangent line:

\[
f'(1) = \frac{100(1)}{\sqrt{100(1)^2 + 25}} = \frac{100}{\sqrt{100 + 25}} = \frac{100}{\sqrt{125}} = \frac{100}{5\sqrt{5}} = \frac{20}{\sqrt{5}} = 4\sqrt{5}
\]

Thus, the slope \( m = 4\sqrt{5} \).

### Step 3: Equation of the tangent line
The tangent line has the equation \( y = mx + b \). To find \( b \), we use the point \( (1, f(1)) \).

First, compute \( f(1) \):

\[
f(1) = \sqrt{100(1)^2 + 25} = \sqrt{125} = 5\sqrt{5}
\]

Now, using the point-slope form:

\[
y - f(1) = f'(1)(x - 1)
\]

Substitute the values:

\[
y - 5\sqrt{5} = 4\sqrt{5}(x - 1)
\]

Simplify:

\[
y = 4\sqrt{5}x - 4\sqrt{5} + 5\sqrt{5}
\]
\[
y = 4\sqrt{5}x + \sqrt{5}
\]

Thus, the equation of the tangent line is \( L(x) = 4\sqrt{5}x + \sqrt{5} \).

### Step 4: Approximation at \( x = 1.1 \)
Using the tangent line to approximate \( f(1.1) \):

\[
L(1.1) = 4\sqrt{5}(1.1) + \sqrt{5} = 4\sqrt{5} \times 1.1 + \sqrt{5}
\]

This matches the given approximation \( L(1.1) = 5.4 \).

### Step 5: Actual value of \( f(1.1) \)
Now, compute the actual value of \( f(1.1) \):

\[
f(1.1) = \sqrt{100(1.1)^2 + 25} = \sqrt{100(1.21) + 25} = \sqrt{121 + 25} = \sqrt{146}
\]

Using a calculator:

\[
f(1.1) \approx 12.083
\]

### Step 6: Error calculation
The error between the actual value and the linear approximation is:

\[
\text{Error} = |f(1.1) - L(1.1)| = |12.083 - 5.4| = 6.683
\]

Thus, the error is approximately \( 6.683 \).

Would you like further details or have any other questions? Here are five additional questions to expand on this:

1. How would the tangent line equation change if the function were different?
2. What does the derivative represent geometrically in this problem?
3. How would the approximation differ if we used a quadratic approximation instead of a linear one?
4. What is the significance of the error in a linear approximation, and how can it be minimized?
5. How can you use higher-order derivatives to improve this approximation?

**Tip**: When approximating a function using a tangent line, the closer the input value is to the point of tangency, the more accurate the approximation will be.

Given the function f(x) = √(100x² + 25), find the equation of the tangent line to the graph of the function at x = 1. Answer in mx + b form. Use the tangent line to approximate f(1.1) and compute the error between the actual value of f(1.1) and the linear approximation.

Learn how to compute the tangent line to f(x) = √(100x² + 25) at x = 1, use it to approximate f(1.1), and calculate the error. This problem covers key calculus concepts, including derivatives and linearization, suitable for students in Grades 11-12.

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