Math Problem Statement

Use linear approximation to estimate \(\sqrt{16.1}\) using the function \(f(x) = \sqrt{x}\) and the tangent line at \(x = 16\). Find the slope (m) and intercept (b) of the tangent line, and use this to approximate \(\sqrt{16.1}\).

Solution

Let's go through the steps to solve the problem visible in the image. The task asks for using linear approximation to estimate 16.1\sqrt{16.1}, using the function f(x)=xf(x) = \sqrt{x} and the tangent line at x=16x = 16.

Step 1: Find the equation of the tangent line at x=16x = 16

The formula for linear approximation (tangent line approximation) is:

L(x)=f(a)+f(a)(xa)L(x) = f(a) + f'(a)(x - a)

where aa is the point where the tangent line is computed (in this case a=16a = 16).

  1. Calculate f(16)f(16): f(16)=16=4f(16) = \sqrt{16} = 4

  2. Find f(x)f'(x): The derivative of f(x)=xf(x) = \sqrt{x} is: f(x)=12xf'(x) = \frac{1}{2\sqrt{x}}

    Now, calculate f(16)f'(16): f(16)=1216=18f'(16) = \frac{1}{2\sqrt{16}} = \frac{1}{8}

    So, the slope m=f(16)=18m = f'(16) = \frac{1}{8}.

Step 2: Write the tangent line equation

The equation of the tangent line at x=16x = 16 is: y=f(16)+f(16)(x16)y = f(16) + f'(16)(x - 16) Substitute the values: y=4+18(x16)y = 4 + \frac{1}{8}(x - 16) This is the equation of the tangent line.

Step 3: Approximate 16.1\sqrt{16.1}

Now, use this tangent line equation to approximate 16.1\sqrt{16.1}, which corresponds to x=16.1x = 16.1.

Substitute x=16.1x = 16.1 into the tangent line equation: y=4+18(16.116)y = 4 + \frac{1}{8}(16.1 - 16) y=4+18(0.1)=4+0.0125=4.0125y = 4 + \frac{1}{8}(0.1) = 4 + 0.0125 = 4.0125

Therefore, the approximation for 16.1\sqrt{16.1} is 4.0125.

Final Answers:

  • Slope m=18=0.125m = \frac{1}{8} = 0.125
  • Intercept b=4b = 4
  • Approximation of 16.1=4.0125\sqrt{16.1} = 4.0125

Would you like further details or have any questions?

Here are five follow-up questions related to this concept:

  1. How does linear approximation differ from other methods of approximation?
  2. Why is the tangent line used to approximate values of the function near a specific point?
  3. How does the accuracy of the approximation change as we move further from the point of tangency?
  4. Could we use linear approximation for non-continuous functions? Why or why not?
  5. How would the approximation change if we used a different point, say x=9x = 9, instead of x=16x = 16?

Tip: Linear approximation is generally more accurate the closer xx 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