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Question 4 In this problem we consider f(x)=√ near x = 16. (a) Find L(r), the linear approximation of f(x) for r near 16. (b) Use L(a) to approximate f(16.1) and f(16.01). (c) Recall that if f is twice differentiable lim 2 f"(a). Therefore, this means that |f" \/" (a) (Ax)², where Az = x-a as usual. With this in mind, what do you think approximately happens to the error when Ar is made ten times smaller? One hundred times smaller? f(x) - L(a) (x-a)² (size of error from approximating f(x)=L(x)) = f(x) - L(x)|≈! (d) Use a calculator to find the errors f(16.1) - L(16.1)| and f(16.01) - L(16.01). Is this consistent with your prediction in the previous part? (e) Find Q(x), the quadratic approximation of f(a) for a near 16. (f) Use Q(x) to approximate f(16.1) and f(16.01). f(x) - Q(x) (g) Use a calculator to find the errors |f(16.1) - Q(16.1)| and f(16.01) - Q(16.01). Note: If f is three times differentiable, then lim Therefore, the size of the quadratic (Ax)³ proportional to Ar]³, with constant related to the third Az-0 6 approximation error is approximately derivative!

Question 4 In this problem we consider f(x)=√ near x = 16. (a) Find L(r), the linear approximation of f(x) for r near 16. (b) Use L(a) to approximate f(16.1) and f(16.01). (c) Recall that if f is twice differentiable lim 2 f"(a). Therefore, this means that |f" \/" (a) (Ax)², where Az = x-a as usual. With this in mind, what do you think approximately happens to the error when Ar is made ten times smaller? One hundred times smaller? f(x) - L(a) (x-a)² (size of error from approximating f(x)=L(x)) = f(x) - L(x)|≈! (d) Use a calculator to find the errors f(16.1) - L(16.1)| and f(16.01) - L(16.01). Is this consistent with your prediction in the previous part? (e) Find Q(x), the quadratic approximation of f(a) for a near 16. (f) Use Q(x) to approximate f(16.1) and f(16.01). f(x) - Q(x) (g) Use a calculator to find the errors |f(16.1) - Q(16.1)| and f(16.01) - Q(16.01). Note: If f is three times differentiable, then lim Therefore, the size of the quadratic (Ax)³ proportional to Ar]³, with constant related to the third Az-0 6 approximation error is approximately derivative!

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter7: Integration
Section7.1: Antiderivatives
Problem 45E
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Question 4 In this problem we consider f(x)=√ near x = 16. (a) Find L(x), the linear approximation of f(x) for x near 16. (b) Use L(a) to approximate f(16.1) and f(16.01). (c) Recall that if f is twice differentiable lim f(x) - L(x) (x-a)² f"(a). Therefore, this means that 2 (size of error from approximating f(x) ≈L(x)) = f(x) — L(x)| ≈ f" (a) (Az)², where Az = x-a as usual. With this in mind, what do you think approximately happens to the error when Ar is made ten times smaller? One hundred times smaller? (d) Use a calculator to find the errors f(16.1) - L(16.1)| and f(16.01) - L(16.01). Is this consistent with your prediction in the previous part? (e) Find Q(x), the quadratic approximation of f(a) for a near 16. (f) Use Q(x) to approximate f(16.1) and f(16.01). (g) Use a calculator to find the errors |f(16.1) - Q(16.1)| and f(16.01) - Q(16.01). Note: If f is three times differentiable, then lim f()-Q(a) f(a). Therefore, the size of the quadratic Az-0 (Ax)³ 6 approximation error is approximately proportional to Az³, with constant related to the third derivative!
Transcribed Image Text:Question 4 In this problem we consider f(x)=√ near x = 16. (a) Find L(x), the linear approximation of f(x) for x near 16. (b) Use L(a) to approximate f(16.1) and f(16.01). (c) Recall that if f is twice differentiable lim f(x) - L(x) (x-a)² f"(a). Therefore, this means that 2 (size of error from approximating f(x) ≈L(x)) = f(x) — L(x)| ≈ f" (a) (Az)², where Az = x-a as usual. With this in mind, what do you think approximately happens to the error when Ar is made ten times smaller? One hundred times smaller? (d) Use a calculator to find the errors f(16.1) - L(16.1)| and f(16.01) - L(16.01). Is this consistent with your prediction in the previous part? (e) Find Q(x), the quadratic approximation of f(a) for a near 16. (f) Use Q(x) to approximate f(16.1) and f(16.01). (g) Use a calculator to find the errors |f(16.1) - Q(16.1)| and f(16.01) - Q(16.01). Note: If f is three times differentiable, then lim f()-Q(a) f(a). Therefore, the size of the quadratic Az-0 (Ax)³ 6 approximation error is approximately proportional to Az³, with constant related to the third derivative!
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