Math Problem Statement
Solution
Let's go through the steps to solve the problem visible in the image. The task asks for using linear approximation to estimate , using the function and the tangent line at .
Step 1: Find the equation of the tangent line at
The formula for linear approximation (tangent line approximation) is:
where is the point where the tangent line is computed (in this case ).
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Calculate :
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Find : The derivative of is:
Now, calculate :
So, the slope .
Step 2: Write the tangent line equation
The equation of the tangent line at is: Substitute the values: This is the equation of the tangent line.
Step 3: Approximate
Now, use this tangent line equation to approximate , which corresponds to .
Substitute into the tangent line equation:
Therefore, the approximation for is 4.0125.
Final Answers:
- Slope
- Intercept
- Approximation of
Would you like further details or have any questions?
Here are five follow-up questions related to this concept:
- How does linear approximation differ from other methods of approximation?
- Why is the tangent line used to approximate values of the function near a specific point?
- How does the accuracy of the approximation change as we move further from the point of tangency?
- Could we use linear approximation for non-continuous functions? Why or why not?
- How would the approximation change if we used a different point, say , instead of ?
Tip: Linear approximation is generally more accurate the closer is to the point of tangency.
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Math Problem Analysis
Mathematical Concepts
Calculus
Linear Approximation
Derivatives
Formulas
Linear approximation formula: L(x) = f(a) + f'(a)(x - a)
Derivative of \(\sqrt{x}\): f'(x) = \frac{1}{2\sqrt{x}}
Theorems
Tangent Line Approximation
Suitable Grade Level
Grades 11-12, College