Approximation Error

In Exerc...

Question

Approximation Error

In Exercises 29–34, each function f(x) changes value when x changes from x₀ to x₀ + dx. Find

a. the change Δf = f(x₀ + dx) − f(x₀);

b. the value of the estimate df = fʹ(x₀) dx; and

c. the approximation error |Δf − df|.

f(x) = x² + 2x, x₀ = 1, dx = 0.1

Accepted Answer

Welcome back everyone. Consider the function H of Z equals 3C2 minus 8z where Z varies from Z to Z + D Z. Determine the following number 1, the change in H denoted as delta H equals h of Z0 plus Dz minus h of Z02 the value of the approximation DH equals. H C not multiplied by Dz and 3 the error in the approximation deltah minus DH calculated within the absolute value function. Let C not be equal to 1 and D Z be equal to 0.05. So let's begin the calculations. Starting with part one, we want to calculate the change in age. We know that delta H is going to be H at C not plus Dz we're given those, so we're calculating H at 1 + 0.05 minus H at C0. So minus H at 1. Simplifying, we can show that we're evaluating H at 1.05. Minus H at 1. Now we can substitute those values into the function, which basically gives us 3 multiplied by 1.05 squared. Minus 8 multiplied by Z. In this case it is 1.05. So this is our H at 1.05, and we have to subtract H at 1. So 3 multiplied by 1 squared minus 8 multiplied by 1. And when we perform the calculations, we essentially arrive at -0.0925, which is the answer for the first parts we can label it as our final answer. Now moving on to the second part, we want to calculate the value of the approximation DH. equals h at Z not multiplied by DC. In this context, we understand that we have to identify the derivative H of C. So let's differentiate the function 3C2 minus 8C. Here we can apply the power rule. So we get 3 multiplied by 2C, which is 6C. -8 multiplied by c derivative, which is just 8. And now let's identify H at Cno, which is H at 1. We get 6 multiplied by 1 minus 8. Which is -2, right? And that said, we can essentially use our formula and show that DH equals H at 1, which is neg2, multiplied by DZ 0.05. When we perform. The calculation, we end up with -0.1. Which is our second answer, right? Let's label it. And finally, we can estimate the error in the approximation, the absolute value of delta H minus DH. So let's substitute the values and we know from the first part that delta H is -0.0925. We are subtracting DH, which is 0.1. And we're taking the absolute value of the result. So this gives us the absolute value of 0.1. We can take the positive number because we have minus a negative number. -0.0925. Calculating the result, we end up with 0.0075, which is our third answer to this problem. Well done. Thank you for watching.

Key Concepts

Function Change Δf

Differential Estimate df

Approximation Error |Δf − df|

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Approximation ErrorIn Exercises 29–34, each function f(x) changes value when x changes from x₀ to x₀ + dx. Finda. the change Δf = f(x₀ + dx) − f(x₀);b. the value of the estimate df = fʹ(x₀) dx; andc. the approximation error |Δf − df|.<IMAGE>f(x) = x² + 2x, x₀ = 1, dx = 0.1

Key Concepts

Function Change Δf

Differential Estimate df

Approximation Error |Δf − df|

Watch next