a)Approximate root 26 by using the linearization of y= root x at the point (25,5) Show the computation that leads to your conclusion.
b)Approximate cube root of 26 by using an appropriate linearization. Show the computation that leads to your conclusion.
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a) To approximate the square root of 26 using the linearization of y = √x at the point (25, 5), we can follow these steps:
Step 1: Write down the equation of the linearization. The equation of the linearization near the point (25, 5) can be given by:
y ≈ f(a) + f'(a) * (x - a)
In this case, a = 25 and f(x) = √x. Therefore, the equation becomes:
y ≈ √25 + (1/(2√25)) * (x - 25)
Step 2: Substitute x = 26 into the equation of the linearization:
y ≈ √25 + (1/(2√25)) * (26 - 25)
Step 3: Simplify the expression:
y ≈ 5 + (1/10) * 1
y ≈ 5.1
Therefore, the approximate square root of 26 using the linearization of y = √x at the point (25, 5) is around 5.1.
b) To approximate the cube root of 26 using an appropriate linearization, we can follow these steps:
Step 1: Write down the equation of the linearization. The equation of the linearization near the point (27, 3) can be given by:
y ≈ f(a) + f'(a) * (x - a)
In this case, a = 27 and f(x) = x^(1/3). Therefore, the equation becomes:
y ≈ 27^(1/3) + (1/3)*(27^(-2/3)) * (x - 27)
Step 2: Substitute x = 26 into the equation of the linearization:
y ≈ 27^(1/3) + (1/3)*(27^(-2/3)) * (26 - 27)
Step 3: Simplify the expression:
y ≈ 3 + (1/3)*(27^(-2/3)) * (-1)
y ≈ 3 - (1/3)*(1/3)*(1/9)
y ≈ 3 - 1/81
y ≈ 2.987
Therefore, the approximate cube root of 26 using the linearization of y = x^(1/3) at the point (27, 3) is around 2.987.