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!
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Question 4** In this problem we consider \( f(x) = \sqrt[4]{x} \) near \( x = 16 \).
(a) Find \( L(x) \), the linear approximation of \( f(x) \) for \( x \) near 16.
(b) Use \( L(x) \) to approximate \( f(16.1) \) and \( f(16.01) \).
(c) Recall that if \( f \) is twice differentiable, then
\[
\lim_{x \to a} \frac{f(x) - L(x)}{(x - a)^2} = \frac{f''(a)}{2}.
\]
Therefore, this means that
\[
\text{(size of error from approximating } f(x) \approx L(x)) = |f(x) - L(x)| \approx \frac{|f''(a)|}{2} (\Delta x)^2,
\]
where \( \Delta x = x - a \) as usual. With this in mind, what do you think approximately happens to the error when \( \Delta x \) 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 \( f(x) \) for \( x \) 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
\[
\lim_{\Delta x \to 0} \frac{f(x) - Q(x)}{(\Delta x)^3} = \frac{f'''(a)}{6}.
\]
Therefore, the size of the quadratic approximation error is approximately proportional to \( |\Delta x|^3 \), with constant related to the third derivative!](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F832d3022-50d6-485d-bc6c-8a554d15f25c%2F353e74cf-4d5b-424b-8ba6-9fb3e93c6110%2Fyp0103l_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 4** In this problem we consider \( f(x) = \sqrt[4]{x} \) near \( x = 16 \).
(a) Find \( L(x) \), the linear approximation of \( f(x) \) for \( x \) near 16.
(b) Use \( L(x) \) to approximate \( f(16.1) \) and \( f(16.01) \).
(c) Recall that if \( f \) is twice differentiable, then
\[
\lim_{x \to a} \frac{f(x) - L(x)}{(x - a)^2} = \frac{f''(a)}{2}.
\]
Therefore, this means that
\[
\text{(size of error from approximating } f(x) \approx L(x)) = |f(x) - L(x)| \approx \frac{|f''(a)|}{2} (\Delta x)^2,
\]
where \( \Delta x = x - a \) as usual. With this in mind, what do you think approximately happens to the error when \( \Delta x \) 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 \( f(x) \) for \( x \) 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
\[
\lim_{\Delta x \to 0} \frac{f(x) - Q(x)}{(\Delta x)^3} = \frac{f'''(a)}{6}.
\]
Therefore, the size of the quadratic approximation error is approximately proportional to \( |\Delta x|^3 \), with constant related to the third derivative!
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