MATH241 Lecture Notes - Lecture 22: Hyperbola, Even And Odd Functions, Quotient Rule
9 May 2018
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MATH241 - Lecture 22 - Linear Approximations and Hyperbolic Functions
3.10: Linear Approximations
The tangent line to a function at , , can be used toy=f(x)x=a y =f(a) + f′(a) (x)− a
approximate or estimate the value for when is near f(x)x a
y−y1=m(x)− x1
→ another way of expressing the same thingy=y1+m(x)− x1
For values of close to x a f (x) ≈ f(a) + f′(a) (x)− a
The linear approximation of at is f x =a L (x) = f(a) + f′(a) (x)− a
Example
:
A) Find the linear approximation of at L(x)xf (x) = x3+ 2 x= 2
B) Use to estimate L(x)f(1.98)
Solution:
A) 2f(2) = 23+ 2 (2) = 1
f′(x) = 3x2+ 2
(2) 4f′(2) = 3 2+ 2 = 1
L(x) = f(2) + f′(2) (x)− 2
2 4= 1 + 1 (x)− 2
B) 2 4L(1.98) = 1 + 1 (1.98 )− 2
2 4= 1 + 1 (− .02)0
2 .28= 1 − 0
1.72= 1
For comparison: 1.7223f(1.98) ≈ 1
Example
:
Approximate using linear approximation√66
Solution:
Choose with f(x) = √x4a= 6
f(64) = 8
xf′(x) = 2
1 −2
1
=1
2√x
f′(64) = 1
2√64
=1
2.8 =1
16
Thus L(x) = f(a) + f′(a) (x)− a
= 8 + 1
16 (x4)− 6
L(66) = 8 + 1
16 (66 4)− 6
= 8 + 1
16 (2)
= 8 + 2
16
= 8 + 8
1 , 25= 8 1
For comparison: , 240
√66 ≈ 8 1
Example
:
Find the linear approximation of at . Use it to estimate in xf (x) = s x = 0 in s(5
1)
Solution:
inf (0) = s(0) = 0
os xf′(x) = c
osf′(0) = c(0) = 1
L(x) = f(a) + f′(a) (x)− a
= 0 + 1 (x)− 0 = x
Document Summary
Math241 - lecture 22 - linear approximations and hyperbolic functions. The tangent line to a function y = f (x) approximate or estimate the value for at f (x) x = a y = f (a) + f (a) (x. , can be used to when x is near a y y1 = m (x y = y1 + m (x. Another way of expressing the same thing. For values of x close to a f (x) f (a) + f (a) (x. The linear approximation of f at x = a is. Example : find the linear approximation f (1. 98) to estimate, use. L (x) of f (x) = x3 + 2 x at x = 2. ) 2 f (2) = 23 + 2 (2) = 1. 2 f (x) = 3x2 + 2. 2 + 2 = 1 f (2) = 3. L (x) = f (2) + f (2) (x.
