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Worksheet: Linear Approximation

Q1:

Find the linear approximation of the function at .

  • A
  • B
  • C
  • D
  • E

Q2:

Find the linear approximation of the function 𝑓 ( π‘₯ ) = √ π‘₯ at π‘₯ = 4 .

  • A 1 2 π‘₯
  • B 1 4 π‘₯ βˆ’ 1
  • C 1 4 π‘₯ + 2
  • D 1 4 π‘₯ + 1
  • E π‘₯ βˆ’ 2

Q3:

What is the tangent line approximation 𝐿 ( π‘₯ ) of √ 1 βˆ’ π‘₯ near π‘₯ = 0 ?

  • A 1 βˆ’ π‘₯ 2
  • B 1 + π‘₯ 2
  • C 1 + π‘₯ 2
  • D 1 βˆ’ π‘₯ 2
  • E 1 βˆ’ √ π‘₯ 2

Q4:

Find the linear approximation of the function 𝑓 ( π‘₯ ) = √ π‘₯ 3 at π‘₯ = βˆ’ 8 .

  • A 1 4 π‘₯ βˆ’ 4
  • B 1 1 2 π‘₯ + 2 3
  • C 1 1 2 π‘₯ βˆ’ 2
  • D 1 1 2 π‘₯ βˆ’ 4 3
  • E βˆ’ 2 π‘₯ βˆ’ 1 9 1 1 2

Q5:

Find the linear approximation of the function 𝑓 ( π‘₯ ) = π‘₯ π‘₯ + 1 at π‘₯ = 1 .

  • A 1 2 π‘₯
  • B 1 4 π‘₯ βˆ’ 1 4
  • C 1 4 π‘₯ + 1 2
  • D 1 4 π‘₯ + 1 4
  • E 1 2 π‘₯ βˆ’ 1 4

Q6:

Find the linear approximation of the function 𝑓 ( π‘₯ ) = π‘₯ s i n at π‘₯ = πœ‹ 6 .

  • A βˆ’ √ 3 2 π‘₯ + √ 3 πœ‹ 1 2 + 1 2
  • B √ 3 2 π‘₯ βˆ’ √ 3 πœ‹ 1 2
  • C √ 3 2 π‘₯ + 1 2
  • D √ 3 2 π‘₯ βˆ’ √ 3 πœ‹ 1 2 + 1 2
  • E π‘₯ βˆ’ πœ‹ 6 + 1 2

Q7:

Find the linear approximation of the function 𝑓 ( π‘₯ ) = π‘₯ π‘₯ s i n at π‘₯ = 2 πœ‹ .

  • A βˆ’ 2 πœ‹ π‘₯ + 4 πœ‹ 2
  • B 2 πœ‹ π‘₯
  • C 2 πœ‹ π‘₯ + 4 πœ‹ 2
  • D 2 πœ‹ π‘₯ βˆ’ 4 πœ‹ 2
  • E π‘₯ βˆ’ 2 πœ‹

Q8:

Find the linear approximation of the function 𝑓 ( π‘₯ ) = π‘₯ s i n 2 at π‘₯ = 0 .

Q9:

Find the linear approximation of the function 𝑓 ( π‘₯ ) = π‘₯ t a n at π‘₯ = πœ‹ .

  • A πœ‹ βˆ’ π‘₯
  • B π‘₯
  • C π‘₯ + πœ‹
  • D π‘₯ βˆ’ πœ‹
  • E0

Q10:

Find the linear approximation of the function 𝑓 ( π‘₯ ) = 2 π‘₯ at π‘₯ = 0 .

  • A π‘₯ + 1
  • B π‘₯ 2 l n
  • C π‘₯ 2 + 1 l n
  • D π‘₯ 2 + 1 l n
  • E π‘₯ + 2 l n

Q11:

Find the linear approximation of the function 𝑓 ( π‘₯ ) = ( 1 + π‘₯ ) π‘˜ at π‘₯ = 0 .

  • A π‘₯ + π‘˜
  • B π‘˜ π‘₯
  • C 1 βˆ’ π‘˜ π‘₯
  • D π‘˜ π‘₯ + 1
  • E π‘₯ + 1

Q12:

Find the linear approximation of the function 𝑓 ( π‘₯ ) = π‘₯ s i n βˆ’ 1 at π‘₯ = 0 .

  • A π‘₯ + 2 πœ‹
  • B π‘₯ + πœ‹
  • C βˆ’ π‘₯
  • D π‘₯
  • E π‘₯ + πœ‹ 2

Q13:

By finding the linear approximation of the function 𝑓 ( π‘₯ ) = π‘₯ 4 at a suitable value of π‘₯ , estimate the value of ( 1 . 9 9 9 ) 4 .

  • A15.984
  • B15.992
  • C16.032
  • D15.968
  • E16.016

Q14:

By finding the linear approximation of the function 𝑓 ( π‘₯ ) = √ π‘₯ at a suitable value of π‘₯ , estimate the value of √ 1 0 0 . 5 .

Q15:

By finding the linear approximation of the function 𝑓 ( π‘₯ ) = 𝑒 π‘₯ at a suitable value of π‘₯ , estimate the value of 𝑒 0 . 1 .

Q16:

By finding the linear approximation of the function 𝑓 ( π‘₯ ) = √ π‘₯ 3 at a suitable value of π‘₯ , estimate the value of 3 √ 1 0 0 1 .

  • A 3 0 1 3 0
  • B 1 0 0 1 1 0 0
  • C 2 9 9 9 3 0 0
  • D 3 0 0 1 3 0 0
  • E 2 9 9 3 0

Q17:

We will explore why we can call the tangent line approximation the β€œbest” local linearization.

What is the tangent line approximation at π‘₯ = πœ‹ for the function 𝑓 ( π‘₯ ) = 3 . 8 ( π‘₯ ) s i n ?

  • A 𝐿 ( π‘₯ ) = π‘₯ βˆ’ πœ‹
  • B 𝐿 ( π‘₯ ) = βˆ’ 3 . 8
  • C 𝐿 ( π‘₯ ) = 3 . 8 ( π‘₯ βˆ’ πœ‹ )
  • D 𝐿 ( π‘₯ ) = βˆ’ 3 . 8 ( π‘₯ βˆ’ πœ‹ )
  • E 𝐿 ( π‘₯ ) = βˆ’ 3 . 8 ( π‘₯ + πœ‹ )

Suppose 𝐿 ( π‘₯ ) = π‘˜ ( π‘₯ βˆ’ πœ‹ ) is used as a local linearization at π‘₯ = πœ‹ of 𝑓 ( π‘₯ ) = 3 . 8 ( π‘₯ ) s i n . Write the expression for the error 𝐸 ( π‘₯ ) .

  • A 𝐸 ( π‘₯ ) = 3 . 8 ( π‘₯ ) βˆ’ π‘˜ ( π‘₯ βˆ’ πœ‹ ) s i n
  • B 𝐸 ( π‘₯ ) = 3 . 8 ( π‘₯ βˆ’ πœ‹ ) βˆ’ π‘˜ ( π‘₯ βˆ’ πœ‹ ) s i n
  • C 𝐸 ( π‘₯ ) = π‘˜ ( π‘₯ βˆ’ πœ‹ )
  • D 𝐸 ( π‘₯ ) = π‘˜ ( π‘₯ βˆ’ πœ‹ ) βˆ’ 3 . 8 ( π‘₯ ) s i n
  • E 𝐸 ( π‘₯ ) = 3 . 8 ( π‘₯ βˆ’ πœ‹ ) + π‘˜ ( π‘₯ βˆ’ πœ‹ ) s i n

Determine the value of π‘˜ for which

  • A βˆ’ 1
  • B3.8
  • C βˆ’ 3 . 8
  • D0
  • E1

Suppose that 𝑓 is a function that is differentiable at π‘₯ = π‘Ž . Using the local linearization at π‘₯ = π‘Ž given by 𝑓 ( π‘₯ ) β‰ˆ 𝑓 ( π‘Ž ) + π‘˜ ( π‘₯ βˆ’ π‘Ž ) , determine

  • A 𝑓 β€² ( π‘Ž ) βˆ’ π‘˜
  • B 𝑓 ( π‘Ž ) βˆ’ π‘˜
  • C π‘˜
  • D 𝑓 β€² ( π‘Ž )
  • E 𝑓 ( π‘Ž ) + π‘˜

Q18:

By finding the linear approximation of the function 𝑓 ( π‘₯ ) = π‘₯ c o s at a suitable value of π‘₯ , estimate the value of c o s 2 9 ∘ .

  • A 1 2 βˆ’ √ 3 πœ‹ 3 6 0
  • B2
  • C √ 3 2 βˆ’ πœ‹ 3 6 0
  • D √ 3 2 + πœ‹ 3 6 0
  • E 1 2 + √ 3 πœ‹ 3 6 0

Q19:

By finding the linear approximation of the function 𝑓 ( π‘₯ ) = π‘₯ s i n at a suitable value of π‘₯ , estimate the value of s i n ( 3 . 1 4 ) .

  • A0
  • B βˆ’ 1
  • C πœ‹ + 3 . 1 4
  • D πœ‹ βˆ’ 3 . 1 4
  • E 3 . 1 4 βˆ’ πœ‹

Q20:

Use a local linearization near πœ‹ 2 to estimate 𝑓 ο€» √ 2  to 3 decimal places, where 𝑓 ( π‘₯ ) = 2 π‘₯ s i n .

Q21:

By finding the linear approximation of the function 𝑓 ( π‘₯ ) = 1 π‘₯ at a suitable value of π‘₯ , estimate the value of 1 4 . 0 0 2 .

Q22:

Find the local linearization of 1 √ 1 + π‘₯ near π‘₯ = 0 .

  • A 1 + π‘₯ 2
  • B 1 βˆ’ π‘₯
  • C 1 + π‘₯
  • D 1 βˆ’ π‘₯ 2
  • E 1 + π‘₯ 2

Q23:

The first condition for a linear function 𝐿 ( π‘₯ ) to be an approximation to 𝑓 ( π‘₯ ) near π‘₯ = π‘Ž is that 𝐿 ( π‘Ž ) = 𝑓 ( π‘Ž ) . So 𝐿 ( π‘₯ ) = 𝑓 ( π‘Ž ) + π‘š ( π‘₯ βˆ’ π‘Ž ) for some constant π‘š . The error in using 𝐿 instead of 𝑓 at a point π‘₯ is the function 𝐸 ( π‘₯ ) = 𝑓 ( π‘₯ ) βˆ’ 𝐿 ( π‘₯ ) .

The tangent line approximation to 𝑓 near π‘₯ = π‘Ž is the linear approximation 𝐿 ( π‘₯ ) = 𝑓 ( π‘Ž ) + 𝑓 β€² ( π‘Ž ) ( π‘₯ βˆ’ π‘Ž ) . What is the tangent line approximation to 𝑓 ( π‘₯ ) = 𝑒 π‘˜ π‘₯ near π‘₯ = 0 ?

  • A 𝐿 ( π‘₯ ) = 1
  • B 𝐿 ( π‘₯ ) = 1 + π‘₯
  • C 𝐿 ( π‘₯ ) = π‘˜ + π‘₯
  • D 𝐿 ( π‘₯ ) = 1 + π‘˜ π‘₯
  • E 𝐿 ( π‘₯ ) = π‘˜ π‘₯

What is the error in using the tangent line approximation to 𝑓 ( π‘₯ ) = 𝑒 4 . 5 π‘₯ near π‘₯ = 0 at the point 0.1? Give your answer to 5 decimal places.

What is the error in using the tangent line approximation to 𝑓 ( π‘₯ ) = 𝑒 4 . 5 π‘₯ near π‘₯ = 0 at the point 0.01? Give your answer to 5 decimal places.

What is the error in using the tangent line approximation to 𝑓 ( π‘₯ ) = 𝑒 4 . 5 π‘₯ near π‘₯ = 0 at the point 0.001? Give your answer to 5 decimal places.

Q24:

The curve 𝐢 of equation 𝑦 = 𝑒 π‘₯ is concave up everywhere. The line 𝑦 = 0 . 1 2 π‘₯ + 1 . 0 3 is above 𝐢 when π‘₯ = 0 but then must quickly cross 𝐢 at some 𝛿 > 0 near 0.

What equation can we use to find 𝛿 ?

  • A 𝑒 = 0 . 1 2 𝛿 𝛿
  • B 𝑒 = 𝛿 𝛿
  • C 𝑒 = 1 . 0 3 𝛿
  • D 𝑒 = 0 . 1 2 𝛿 + 1 . 0 3 𝛿
  • E 𝑒 = 0 . 1 2 𝛿 βˆ’ 1 . 0 3 𝛿

Find an estimate of 𝛿 using the tangent line approximation of 𝑓 ( π‘₯ ) = 𝑒 π‘₯ at π‘₯ = 0 in the equation above. Give your answer to 3 decimal places.

Q25:

In the figure, which is the graph of the tangent line approximation of 𝑓 ( π‘₯ ) = 𝑒 βˆ’ 2 ( π‘₯ + 1 ) π‘₯ = 1 2 π‘₯ 3 n e a r ?

  • A(b)
  • B(a)
  • C(c)
  • D(d)
  • Enone of the lines shown

Q26:

Suppose that 𝐿 is any linear approximation to the function 𝑓 near π‘₯ = π‘Ž . The error in using 𝐿 instead of 𝑓 at a point π‘₯ is given by 𝐸 ( π‘₯ ) = 𝑓 ( π‘₯ ) βˆ’ 𝐿 ( π‘₯ ) . What is the error function in using the tangent line approximation for 𝑓 ( π‘₯ ) = π‘₯ βˆ’ 3 π‘₯ + π‘₯ + 5 3 2 near π‘₯ = 1 ?

  • A 𝐸 ( π‘₯ ) = π‘₯ βˆ’ 3 π‘₯ + 5 π‘₯ βˆ’ 7 3 2
  • B 𝐸 ( π‘₯ ) = βˆ’ π‘₯ + 3 π‘₯ βˆ’ 3 π‘₯ + 1 3 2
  • C 𝐸 ( π‘₯ ) = π‘₯ βˆ’ 3 π‘₯ βˆ’ π‘₯ + 1 1 3 2
  • D 𝐸 ( π‘₯ ) = π‘₯ βˆ’ 3 π‘₯ + 3 π‘₯ βˆ’ 1 3 2
  • E 𝐸 ( π‘₯ ) = βˆ’ π‘₯ + 3 π‘₯ + π‘₯ βˆ’ 1 1 3 2

Q27:

Let 𝐿 ( π‘₯ ) be the tangent line approximation of the function √ π‘₯ at π‘₯ = 1 , so that 𝐸 ( π‘₯ ) = √ π‘₯ βˆ’ 𝐿 ( π‘₯ ) is its error function. Since the function 𝑔 ( π‘₯ ) = 𝐸 ( π‘₯ ) π‘₯ βˆ’ 1 β†’ 0 as π‘₯ β†’ 1 , we can locally linearize it by π‘˜ ( π‘₯ βˆ’ 1 ) for some constant π‘˜ at π‘₯ = 1 . This means that 𝑔 ( π‘₯ ) β‰ˆ π‘˜ ( π‘₯ βˆ’ 1 ) or 𝑔 ( π‘₯ ) ( π‘₯ βˆ’ 1 ) β‰ˆ π‘˜ . Estimate π‘˜ to 3 decimal places by examining a suitable sequence of values.

Q28:

Suppose that we call the linear function 𝐿 ( π‘₯ ) = 𝑝 + π‘ž π‘₯ a good approximation to 𝑓 near π‘₯ = π‘Ž provided that

What does this tell us about 𝑝 ?

  • A 𝑝 = 0
  • B 𝑝 = 𝑓 ( π‘Ž )
  • C There is nothing we can deduce about the value of 𝑝 .
  • D 𝑝 = 𝑓 ( π‘Ž ) βˆ’ π‘ž π‘Ž

So, we can rewrite 𝐿 ( π‘₯ ) = 𝑓 ( π‘Ž ) + π‘ž ( π‘₯ βˆ’ π‘Ž ) . Suppose that we have another good linear approximation to 𝑓 𝑀 ( π‘₯ ) = 𝑓 ( π‘Ž ) + π‘Ÿ ( π‘₯ βˆ’ π‘Ž ) . How is π‘ž related to π‘Ÿ ?

  • A π‘ž = π‘Ÿ
  • B π‘ž = βˆ’ π‘Ÿ + π‘Ž
  • C π‘ž = π‘Ÿ + π‘Ž
  • D π‘ž = βˆ’ π‘Ÿ
  • E π‘ž = π‘Ÿ π‘Ž

Suppose 𝑓 and 𝑔 are differentiable at π‘₯ = π‘Ž . Then their tangent line approximations 𝑓 ( π‘Ž ) + 𝑓 β€² ( π‘Ž ) ( π‘₯ βˆ’ π‘Ž ) and 𝑔 ( π‘Ž ) + 𝑔 β€² ( π‘Ž ) ( π‘₯ βˆ’ π‘Ž ) are both good approximations near π‘₯ = π‘Ž . Use these two to find a good linear approximation for the product function 𝑓 𝑔 near π‘₯ = π‘Ž in the form 𝐿 ( π‘₯ ) = 𝑓 ( π‘Ž ) 𝑔 ( π‘Ž ) + π‘ž ( π‘₯ βˆ’ π‘Ž ) .

  • A 𝐿 ( π‘₯ ) = 𝑓 ( π‘Ž ) 𝑔 ( π‘Ž ) βˆ’ 𝑓 β€² ( π‘Ž ) 𝑔 β€² ( π‘Ž ) ( π‘₯ βˆ’ π‘Ž )
  • B 𝐿 ( π‘₯ ) = 𝑓 ( π‘Ž ) 𝑔 ( π‘Ž ) + 𝑓 β€² ( π‘Ž ) 𝑔 β€² ( π‘Ž ) ( π‘₯ βˆ’ π‘Ž )
  • C 𝐿 ( π‘₯ ) = 𝑓 ( π‘Ž ) 𝑔 ( π‘Ž ) + ( 𝑓 β€² ( π‘Ž ) 𝑔 ( π‘Ž ) + 𝑓 ( π‘Ž ) 𝑔 β€² ( π‘Ž ) ) ( π‘₯ βˆ’ π‘Ž )
  • D 𝐿 ( π‘₯ ) = ( 𝑓 ( π‘Ž ) 𝑔 ( π‘Ž ) + 𝑓 β€² ( π‘Ž ) 𝑔 β€² ( π‘Ž ) ) ( π‘₯ βˆ’ π‘Ž )
  • E 𝐿 ( π‘₯ ) = ( 𝑓 ( π‘Ž ) 𝑔 ( π‘Ž ) + 𝑓 β€² ( π‘Ž ) 𝑔 ( π‘Ž ) + 𝑓 ( π‘Ž ) 𝑔 β€² ( π‘Ž ) ) ( π‘₯ βˆ’ π‘Ž )

What can you conclude about the derivative ( 𝑓 𝑔 ) β€² ( π‘Ž ) ?

  • A ( 𝑓 𝑔 ) β€² ( π‘Ž ) = 𝑓 β€² ( π‘Ž ) 𝑔 ( π‘Ž ) + 𝑓 ( π‘Ž ) 𝑔 β€² ( π‘Ž )
  • B ( 𝑓 𝑔 ) β€² ( π‘Ž ) = 𝑓 β€² ( π‘Ž ) 𝑔 ( π‘Ž ) βˆ’ 𝑓 ( π‘Ž ) 𝑔 β€² ( π‘Ž )
  • C ( 𝑓 𝑔 ) β€² ( π‘Ž ) = 𝑓 ( π‘Ž ) 𝑔 ( π‘Ž ) + 𝑓 β€² ( π‘Ž ) 𝑔 β€² ( π‘Ž )
  • D ( 𝑓 𝑔 ) β€² ( π‘Ž ) = 𝑓 ( π‘Ž ) 𝑔 ( π‘Ž ) βˆ’ 𝑓 β€² ( π‘Ž ) 𝑔 β€² ( π‘Ž )

Q29:

Use the approximation ( 1 + π‘₯ ) β‰ˆ 1 + π‘˜ π‘₯ π‘˜ , which is valid for small values of π‘₯ , to estimate ( 1 . 0 0 0 2 ) 5 0 .

  • A0.96
  • B0.99
  • C1.03
  • D1.01
  • E1.02

Q30:

We want to find cube roots using the cube function 𝑓 ( π‘₯ ) = π‘₯ 3 .

The cube root of 7 can be found as the zero of 𝑓 ( π‘₯ ) βˆ’ 7 , which is just π‘₯ βˆ’ 7 3 . What is the local linearization of π‘₯ βˆ’ 7 3 at π‘₯ = 2 ?

  • A 1 2 ( π‘₯ βˆ’ 2 )
  • B 1 + 1 1 2 ( π‘₯ βˆ’ 2 )
  • C1
  • D 1 + 1 2 ( π‘₯ βˆ’ 2 )
  • E 1 + 1 2 π‘₯

By finding the intersection with the π‘₯ -axis, what does the tangent line linearization give as an estimate of 3 √ 7 ? Answer with a fraction. We will call this value 𝑝 .

  • A 2 3 1 2
  • B βˆ’ 1 6
  • C 2 5 1 2
  • D βˆ’ 1 1 2
  • E 1 6

Using the value 𝑝 above, what is the tangent line linearization of π‘₯ βˆ’ 7 3 at π‘₯ = 𝑝 ? Give an exact answer.

  • A 7 1 1 7 2 8 + 2 3 4 ο€Ό π‘₯ βˆ’ 2 3 1 2 
  • B 7 1 1 7 2 8 + 5 2 9 1 4 4 ο€Ό π‘₯ βˆ’ 2 3 1 2 
  • C 7 1 1 7 2 8 + 5 2 9 4 8 ο€Ό π‘₯ βˆ’ 2 3 1 2 
  • D 7 1 1 7 2 8
  • E 7 1 1 7 2 8 + 7 1 1 7 2 8 ο€Ό π‘₯ βˆ’ 2 3 1 2 

Using this new tangent line, give a second estimate of the cube root 3 √ 7 accurate to five decimal places.

  • A1.91294
  • B βˆ’ 1 . 9 2 0 3 9
  • C βˆ’ 1 . 9 1 2 9 4
  • D βˆ’ 0 . 0 0 0 3 7
  • E 1.92039

Q31:

By factoring the error function in using the tangent line approximation to 𝑓 ( π‘₯ ) = 2 π‘₯ + 6 π‘₯ + 7 π‘₯ βˆ’ 5 3 2 near π‘₯ = βˆ’ 1 , or otherwise, find the largest 𝛿 so that if βˆ’ 1 βˆ’ 𝛿 < π‘₯ < βˆ’ 1 + 𝛿 then this error is less than 0.01. Give your 𝛿 to 3 decimal places.

Q32:

The tangent line gives a linear approximation to a function near a point. We consider higher order polynomials.

Suppose 𝑓 is twice differentiable at π‘₯ = π‘Ž . Let 𝑔 ( π‘₯ ) = 𝐴 + 𝐡 ( π‘₯ βˆ’ π‘Ž ) + 𝐢 ( π‘₯ βˆ’ π‘Ž ) 2 be the polynomial satisfying 𝑔 ( π‘Ž ) = 𝑓 ( π‘Ž ) , 𝑔 β€² ( π‘Ž ) = 𝑓 β€² ( π‘Ž ) , and 𝑔 β€² β€² ( π‘Ž ) = 𝑓 β€² β€² ( π‘Ž ) . In terms of 𝑓 , what are the coefficients 𝐴 , 𝐡 , and 𝐢 ?

  • A 𝑓 ( π‘Ž ) , 𝑓 ( π‘Ž ) , 𝑓 β€² ( π‘Ž )
  • B 𝑓 ( π‘Ž ) , 𝑓 β€² ( π‘Ž ) , 𝑓 β€² β€² ( π‘Ž )
  • C 𝑓 β€² ( π‘Ž ) 𝑓 ( π‘Ž ) , 𝑓 β€² ( π‘Ž ) , 𝑓 β€² ( π‘Ž )
  • D 𝑓 ( π‘Ž ) , 𝑓 β€² ( π‘Ž ) , 𝑓 β€² β€² ( π‘Ž ) 2
  • E 𝑓 β€² β€² ( π‘Ž ) 𝑓 ( π‘Ž ) , 𝑓 β€² β€² ( π‘Ž ) , 𝑓 β€² β€² ( π‘Ž )

Is 𝑔 always a quadratic polynomial? Why?

  • A no, because the point π‘Ž may be an inflection point of 𝑓
  • B yes, because we have coefficients up to degree 2

What is 𝑔 ( π‘₯ ) when 𝑓 ( π‘₯ ) = √ π‘₯ 3 at π‘₯ = 8 ? Give your coefficients as fractions.

  • A 1 1 4 4 + 1 2 8 8 ( π‘₯ βˆ’ 8 ) + 1 5 7 6 ( π‘₯ βˆ’ 8 ) 2
  • B 2 + 1 1 2 ( π‘₯ βˆ’ 8 ) βˆ’ 1 1 4 4 ( π‘₯ βˆ’ 8 ) 2
  • C 2 + 1 1 2 ( π‘₯ βˆ’ 8 ) βˆ’ 1 2 8 8 ( π‘₯ βˆ’ 8 ) 2
  • D 2 + 2 ( π‘₯ βˆ’ 8 ) + 1 1 2 ( π‘₯ βˆ’ 8 ) 2
  • E 1 6 + 1 1 2 ( π‘₯ βˆ’ 8 ) + 1 1 2 ( π‘₯ βˆ’ 8 ) 2

Using the function 3 √ π‘₯ at π‘₯ = 8 , find the tangent line approximation of the cube root of 7 to 5 decimal places.

Using the function 3 √ π‘₯ at π‘₯ = 8 , determine the quadratic approximation of the cube root of 7 to 5 decimal places.

Q33:

We consider the local linearization of the natural log function near π‘₯ = 1 .

Find the local linearization of 𝑓 ( π‘₯ ) = π‘₯ l n near π‘₯ = 1 .

  • A 𝐿 ( π‘₯ ) = 1 βˆ’ π‘₯
  • B 𝐿 ( π‘₯ ) = π‘₯
  • C 𝐿 ( π‘₯ ) = π‘₯ + 1
  • D 𝐿 ( π‘₯ ) = π‘₯ βˆ’ 1
  • E 𝐿 ( π‘₯ ) = 1 π‘₯

Knowing now that l i m π‘₯ β†’ 1 𝐸 ( π‘₯ ) π‘₯ βˆ’ 1 = 0 , where 𝐸 ( π‘₯ ) is the error function, what is l i m l n π‘₯ β†’ 1 π‘₯ π‘₯ βˆ’ 1 ?