Questions: Use linear approximation, i.e. the tangent line, to approximate √49.3 as follows: Let f(x)=√x. Find the equation of the tangent line to f(x) at x=49 L(x)= Using this, we find our approximation for √49.3 is

Use linear approximation, i.e. the tangent line, to approximate √49.3 as follows: Let f(x)=√x. Find the equation of the tangent line to f(x) at x=49

L(x)=

Using this, we find our approximation for √49.3 is

Transcript text: Use linear approximation, i.e. the tangent line, to approximate $\sqrt{49.3}$ as follows: Let $f(x)=\sqrt{x}$. Find the equation of the tangent line to $f(x)$ at $x=49$ \[ L(x)= \] Using this, we find our approximation for $\sqrt{49.3}$ is

Solution

Solution Steps

Step 1: Find the derivative of $f(x)$

The derivative of $f(x)$, denoted as $f'(x)$, is found to be $\frac{1}{2 \sqrt{x}}$.

Step 2: Calculate the slope of the tangent line at $x=a$

The slope of the tangent line at $x=49$ is $m = f'(49) = 1/14$.

Step 3: Determine the y-intercept $b$ of the tangent line

Using the point-slope form, the y-intercept $b$ is calculated as $b = f(49) - m \cdot 49 = 7/2$.

Step 4: Form the equation of the tangent line $L(x)$

The equation of the tangent line is $L(x) = 1/14 \cdot (x - 49) + 7 = \frac{x}{14} + \frac{7}{2}$.

Step 5: Approximate $f(a+h)$

By evaluating $L(a+h)$, the linear approximation of $f(x)$ at $x = 49 + 0.3$ is approximately $7.021$.

Final Answer:

The linear approximation of $f(49+0.3)$ is approximately $7.021$, rounded to 9 decimal places.

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