Find the linearization of the function f(x) = x + 6 at a = 3 and use it to approximate the numbers 8.99 and 9.01. Are these approximations overestimates or underestimates?SOLUTION The derivative of f(x) = (x + 6)1/2 isf '(x) = (12)(x+6)(−(12))·1 .and so we have f(3) = and f '(3) = . Putting these values into this equation, we see that the linearization isL(x)= f(3) + f '(3)(x − 3) = + (x − 3)= (16)x+172 .The corresponding linear approximation isx + 6 ≈ + x6 (when x is near 3).In particular, we have8.99 ≈ 52 + 6 = (round to four decimal places)and 9.01 ≈ 52 + 6 = (round to four decimal places).The linear approximation is illustrated in the figure to the left. We see that, indeed, the tangent line approximation is a good approximation to the given function when x is near 3. We also see that our approximations are overestimates because the tangent line lies above the curve. Of course, a calculator could give us approximations for 8.99 and 9.01, but the linear approximation gives an approximation over an entire interval.
Question
Find the linearization of the function f(x) = x + 6 at a = 3 and use it to approximate the numbers 8.99 and 9.01. Are these approximations overestimates or underestimates?SOLUTION The derivative of f(x) = (x + 6)1/2 isf '(x) = (12)(x+6)(−(12))·1 .and so we have f(3) = and f '(3) = . Putting these values into this equation, we see that the linearization isL(x)= f(3) + f '(3)(x − 3) = + (x − 3)= (16)x+172 .The corresponding linear approximation isx + 6 ≈ + x6 (when x is near 3).In particular, we have8.99 ≈ 52 + 6 = (round to four decimal places)and 9.01 ≈ 52 + 6 = (round to four decimal places).The linear approximation is illustrated in the figure to the left. We see that, indeed, the tangent line approximation is a good approximation to the given function when x is near 3. We also see that our approximations are overestimates because the tangent line lies above the curve. Of course, a calculator could give us approximations for 8.99 and 9.01, but the linear approximation gives an approximation over an entire interval.
Solution
I'm sorry, but the text you provided seems to be incomplete and a bit confusing. However, I can help you find the linearization of the function f(x) = x + 6 at a = 3 and use it to approximate the numbers 8.99 and 9.01. Here are the steps:
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First, find the derivative of the function f(x) = x + 6. The derivative, f'(x), is simply 1 because the derivative of x is 1 and the derivative of a constant (6 in this case) is 0.
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Evaluate the function and its derivative at a = 3. We have f(3) = 3 + 6 = 9 and f'(3) = 1.
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The linearization of the function at a = 3 is given by the equation L(x) = f(3) + f'(3)(x - 3). Substituting the values we found in step 2, we get L(x) = 9 + 1*(x - 3) = x + 6.
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Now, we can use this linearization to approximate the numbers 8.99 and 9.01. We have L(8.99) = 8.99 + 6 = 14.99 and L(9.01) = 9.01 + 6 = 15.01.
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To determine whether these approximations are overestimates or underestimates, we can compare them to the actual values of the function at these points. We have f(8.99) = 8.99 + 6 = 14.99 and f(9.01) = 9.01 + 6 = 15.01. Since the approximations are equal to the actual values, they are neither overestimates nor underestimates.
Similar Questions
Find the linearization of the function f(x) = x + 1 at a = 8 and use it to approximate the numbers 8.95 and 9.04. Are these approximations overestimates or underestimates?SolutionThe derivative of f(x) = (x + 1)1⁄2 isf ′(x) = 12(x+1)−(12) ,and so we have f(8) = and f ′(8) = . Putting these values into the equation L(x) = f(a) + f ′(a)(x − a), we see that the linearization isL(x)= f(8) + f '(8)(x − 8) = + (x − 8)= .The corresponding linear approximation isx + 1 ≈ + x6 (when x is near 8).In particular, we have8.95 ≈ 53 + 6 = (round to four decimal places)and 9.04 ≈ 53 + 6 = (round to four decimal places).
The corresponding linear approximation isx + 6 ≈ + x6 (when x is near 3).In particular, we have8.99 ≈ 52 + 6 = (round to four decimal places)and 9.01 ≈ 52 + 6 = (round to four decimal places).The linear approximation is illustrated in the figure to the left. We see that, indeed, the tangent line approximation is a good approximation to the given function when x is near 3. We also see that our approximations are overestimates because the tangent line lies above the curve. Of course, a calculator could give us approximations for 8.99 and 9.01, but the linear approximation gives an approximation over an entire interval.
so we have f(3) = and f '(3) = . Putting these values into this equation, we see that the linearization isL(x)= f(3) + f '(3)(x − 3) = + (x − 3)= .
Find the linearization L(x) of the function at a.f(x) = x3/4, a = 16
Suppose the linear approximation for a function f(x) at a = 2 is given by the tangent line y = −2x + 10.What are f(2) and f '(2)?f(2) = f '(2) = If g(x) =[f(x)]2, find the linear approximation for g(x) at a = 2.L(x) =
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