Determine an appropriate linear approximation of the function f(x)= the square root of x and use it to approximate the square root of 24.2. Write your answer in decimal form
f'=1/2 1/sqrtx
linear approximation: f(24.2)=f(25)-1/2sqrt25*(dx)
= 5-1/2*1/5*.8=5-.08=4.82
Steps above are correct just the answer is wrong because of wrong subtraction. real answer is 4.92
To find an appropriate linear approximation for the function f(x) = √x, we need to choose a point around which we will base our approximation. A commonly used approach is to choose a point near the value we want to approximate.
Let's take a point near x = 25 as our approximation point. The function f(x) = √x can be approximated with the linear function f'(x) = f(a) + f'(a)(x - a), where a is the approximation point.
Since a = 25, we need to find the values of f(25) and f'(25) to continue with the approximation process.
f(a) = f(25) = √25 = 5
f'(x) = (1/2)√x
f'(a) = f'(25) = (1/2)√25 = (1/2)(5) = 5/2
Now we can use the linear approximation to estimate the square root of 24.2 using the point x = 25 as follows:
f'(x) = f(a) + f'(a)(x - a)
f'(24.2) ≈ f(25) + f'(25)(24.2 - 25)
f'(24.2) ≈ 5 + (5/2)(24.2 - 25)
f'(24.2) ≈ 5 + (5/2)(-0.8)
f'(24.2) ≈ 5 + (-2)
f'(24.2) ≈ 3
Therefore, the linear approximation of the square root of 24.2 is approximately 3.
To determine an appropriate linear approximation of the function f(x) = √x, we need to find a linear equation of the form y = mx + b that approximately represents the square root function near the point of interest.
We start by choosing a point near the value we want to approximate. Since we want to approximate √24.2, we can choose a point such as x = 25, which is very close to 24.2.
At x = 25, the value of the square root function is f(25) = √25 = 5.
Now, we need to find the slope of the tangent line to the square root function at x = 25. The slope of the tangent line represents the rate at which the function is changing, which we can find using calculus by taking the derivative of the function.
Using the power rule, the derivative of f(x) = √x is:
f'(x) = (1/2) * x^(-1/2)
Evaluating the derivative at x = 25:
f'(25) = (1/2) * 25^(-1/2) = 1/10
So, the slope of the tangent line at x = 25 is 1/10.
Now, we can substitute the slope and the values of x and y for the point (25, 5) into the equation y = mx + b to solve for b, the y-intercept:
5 = (1/10) * 25 + b
5 = 5/2 + b
b = 5 - 5/2
b = 5/2
Therefore, the linear equation that approximates the square root function near x = 25 is y = (1/10)x + 5/2.
To approximate the square root of 24.2 using this linear approximation, we substitute x = 24.2 into the equation:
y = (1/10) * 24.2 + 5/2
y ≈ 2.42 + 2.5
y ≈ 4.92
Therefore, the square root of 24.2, approximated using the linear approximation, is approximately 4.92 (in decimal form).