Question

use linear approximation, i.e. the tangent line, to approximate \\(\sqrt{36.3}\\) as follows: let \\(f(x) = \sqrt{x}\\). find the equation of the tangent line to \\(f(x)\\) at \\(x = 36\\) \\(l(x) = \square\\) using this, we find our approximation for \\(\sqrt{36.3}\\) is \\(\square\\) note: for this part, give your answer to at least 9 significant figures or use an expression to give the exact answer. question help: video message instructor submit question jump to answer

Explanation:

Step1: Find \( f(36) \) and \( f'(x) \)

Given \( f(x) = \sqrt{x}=x^{\frac{1}{2}} \). First, calculate \( f(36) \): \( f(36)=\sqrt{36} = 6 \). Then, find the derivative \( f'(x) \) using the power rule \( \frac{d}{dx}(x^n)=nx^{n - 1} \). So, \( f'(x)=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}} \).

Step2: Find \( f'(36) \)

Substitute \( x = 36 \) into \( f'(x) \): \( f'(36)=\frac{1}{2\sqrt{36}}=\frac{1}{2\times6}=\frac{1}{12}\approx0.083333333 \).

Step3: Equation of the tangent line \( L(x) \)

The formula for the tangent line (linear approximation) at \( x = a \) is \( L(x)=f(a)+f'(a)(x - a) \). Here, \( a = 36 \), so \( L(x)=f(36)+f'(36)(x - 36) \). Substituting the values we found: \( L(x)=6+\frac{1}{12}(x - 36) \). Simplify this: \( L(x)=6+\frac{1}{12}x-3=\frac{1}{12}x + 3 \).

Step4: Approximate \( \sqrt{36.3} \)

We use \( L(x) \) to approximate \( f(36.3) \), so substitute \( x = 36.3 \) into \( L(x) \): \( L(36.3)=\frac{1}{12}(36.3)+3 \). Calculate \( \frac{36.3}{12}=3.025 \), then \( 3.025 + 3=6.025 \).

Answer:

The equation of the tangent line \( L(x)=\frac{1}{12}x + 3 \) (or \( L(x)=6+\frac{1}{12}(x - 36) \)) and the approximation for \( \sqrt{36.3} \) is \( 6.025 \) (or more precisely, using the exact expression: \( 6+\frac{1}{12}(36.3 - 36)=6+\frac{0.3}{12}=6 + 0.025=6.025 \)). For the tangent line equation, we can also write it as \( L(x)=\frac{1}{12}x+3 \), and the approximation is \( 6.025 \) (with at least 9 significant figures, it can be written as \( 6.02500000 \)).