Fasttasks 2 46 – The Troubleshooting Approximate

1. Fasttasks 2 46 – The Troubleshooting Approximate Error
2. Fasttasks 2 46 – The Troubleshooting Approximate The Number
3. Fasttasks 2 46 – The Troubleshooting Approximate Square

The factor 46.5 is close to 50 and the factor 2.4 is close to 2. Therefore, choosing compatible numbers 50 and 2 would make it easier to multiply mentally. Estimate: Strategy 3: Use compatible numbers to estimate the product. Answer: The estimated product of 46.5 and 2.4 is 100. In Example 2, each factor was changed to a compatible number. Approximate Tile Size: 6x24. For screen reader problems with this website, please call 1-800-430-3376 or text 38698 (standard carrier rates apply to. Documentation for FastTasks 2 (last revised: Oct 31, 2014) Please note we're in the process of updating the documentation for v1.5 and it is not yet complete. In the meantime, please email us if you have specific questions. https://bestaup186.weebly.com/crest-capital-complaints.html. Help on SuperSearch feature can be found here. Item 1: Firewall If the OS X Firewall is. Problem with your Dodge Ram 1500? Our list of 46 known complaints reported by owners can help you fix your Dodge Ram 1500.

Solution.

Here, obviously, (fleft( x right) = {x^{largefrac{2}{3}normalsize}}) and (x = 8,2.) Let ({a = 8}.) Then

[
{fleft( {a = 8} right) = {8^{largefrac{2}{3}normalsize}} }
= {{left( {sqrt[large 3normalsize]{8}} right)^2} = {2^2} = 4.}
]

Find the derivative:

[require{cancel}
{f'left( x right) = {left( {{x^{largefrac{2}{3}normalsize}}} right)^prime } }
= {frac{2}{3}{x^{ – largefrac{1}{3}normalsize}} = frac{2}{{3sqrt[large 3normalsize]{x}}},;;}Rightarrow
{f'left( {a = 8} right) = frac{2}{{3sqrt[large 3normalsize]{8}}} }
= {frac{cancel{2}}{{3 cdot cancel{2}}} = frac{1}{3}.}
]

The result is

[
{fleft( x right) approx fleft( {a} right) + f'left( {a} right)left( {x – a} right),;;}Rightarrow
{{left( {8,2} right)^{largefrac{2}{3}normalsize}} approx 4 + frac{1}{3} cdot left( {8,2 – 8} right) }
= {4 + frac{1}{3} cdot 0,2 approx 4,067.}
]

Solution.

Consider the function (fleft( x right) = {x^n}.) When the independent variable changes by (Delta x), the increment of the function is given by

[Delta y = {left( {x + Delta x} right)^n} – {x^n}.]

If (Delta x) is a small quantity, we can approximately assume that

[
{Delta y approx dy = f'left( x right)Delta x }
= {{left( {{x^n}} right)^prime }Delta x }
= {n{x^{n – 1}}Delta x.}
]

Consequently,

[{left( {x + Delta x} right)^n} approx {x^n} + n{x^{n – 1}}Delta x.]

Suppose further that (x = 1) and (Delta x = alpha.) Then

[
{{left( {1 + alpha } right)^n} }
{approx {1^n} + n cdot {1^{n – 1}} cdot alpha }
= {1 + nalpha .}
]

In particular,

[
{sqrt {1,02} = sqrt {1 + 0,02} }
= {{left( {1 + 0,02} right)^{largefrac{1}{2}normalsize}} }
approx {1 + frac{1}{2} cdot 0,02} = {1,01.}
]

Example 11.

Derive the approximate formulaUsing this identity calculate approximately the value of (sqrt {150} .)

Solution.

Consider the function (y = sqrt x .) If the independent variables changes by (Delta x), the increment of the function is written as

[Delta y = sqrt {x + Delta x} – sqrt x .]

This increment for small (Delta x) can be approximated by the differential so that

[
{Delta y = sqrt {x + Delta x} – sqrt x approx dy }
= {f'left( x right)Delta x }
= {{left( {sqrt x } right)^prime }Delta x }
= {frac{1}{{2sqrt x }}Delta x.}
]

Thus,

Casino payback percentages. [sqrt {x + Delta x} approx sqrt x + frac{1}{{2sqrt x }}Delta x.]

We denote (x = {a^2}), (Delta x = h.) Then we obtain the following approximate equation:

[sqrt {{a^2} + h} approx a + frac{h}{{2a}}.]

Using this formula we can estimate the value of (sqrt {150}:)

[{sqrt {150} = sqrt {144 + 6}} = {sqrt {{{12}^2} + 6}} approx {12 + frac{1}{4} = 12,25.}]

The exact value (up to (3) digits after the decimal point) is equal to (12,247). As it can be seen, the relative error when using the approximate formula is

[
{frac{{12,250 – 12,247}}{{12,247}} = frac{{0,003}}{{12,247}} }
= {2,4 cdot {10^{ – 4}} lt 0,03% .}
]

Solution.

Let (y = sqrt[large nnormalsize]{x}.) If the variable (x) changes by (Delta x,) the increment of the function has the form:

[Delta y = sqrt[large nnormalsize]{{x + Delta x}} – sqrt[large nnormalsize]{x}.]

Considering (Delta x) as a small quantity, we can replace the increment of the function (Delta y) by its differential:

[
{Delta y = sqrt[large nnormalsize]{{x + Delta x}} – sqrt[large nnormalsize]{x} approx dy }
= {f'left( x right)Delta x }
= {{left( {sqrt[large nnormalsize]{x}} right)^prime }Delta x }
= {{left( {{x^{largefrac{1}{n}normalsize}}} right)^prime }Delta x }
= {frac{1}{n}{x^{largefrac{1}{n}normalsize – 1}}Delta x }
= {frac{1}{n}{x^{largefrac{{1 – n}}{n}normalsize}}Delta x.}
]

Then

[sqrt[large nnormalsize]{{x + Delta x}} approx sqrt[large nnormalsize]{x} + frac{1}{n}{x^{largefrac{{1 – n}}{n}normalsize}}Delta x.]

Tumult hype. Denoting (x = {a^n}), (Delta x = h,) we obtain the following relationship:

[
{sqrt[n]{{{a^n} + h}} }
approx {a + frac{1}{n}{left( {{a^n}} right)^{largefrac{{1 – n}}{n}normalsize}}h }
= {a + frac{h}{{n{a^{n – 1}}}}.}
]

Using this formula, we find:

[
{sqrt[large 8normalsize]{{250}} = sqrt[large 8normalsize]{{256 – 6}} }
= {sqrt[large 8normalsize]{{{2^8} – 6}} approx 2 + frac{{left( { – 6} right)}}{{8 cdot {2^7}}} }
= {2 – frac{6}{{1024}} approx 1,994.}
]

Solution.

We choose (a = 45^circ.) The derivative of cosine at this point is equal to

[
{fleft( x right) = cos x,;;}Rightarrow
{f'left( x right) = – sin x,;;}Rightarrow
{f'left( {a = 45^circ} right) }
= { – sin 45^circ = – frac{{sqrt 2 }}{2}.}
]

Express the increment of the independent variable (Delta x) in radians:

[
{Delta x = 46^circ – 45^circ = 1^circ }
= {frac{{2pi }}{{360}} }
= {frac{pi }{{180}};text{radians}.}
]

Using the approximate formula for small (Delta x)

[fleft( x right) approx fleft( {a} right) + f'left( {a} right)Delta x,]

we get:

[
{cos 46^circ approx cos 45^circ + left( { – frac{{sqrt 2 }}{2}} right) cdot frac{pi }{{180}} }
= {frac{{sqrt 2 }}{2} – frac{{sqrt 2 }}{2} cdot frac{pi }{{180}} }
= {frac{{sqrt 2 }}{2}left( {1 – frac{pi }{{180}}} right) }
approx {0,7071 cdot left( {1 – 0,0175} right) }
={ 0,6948.}
]

Solution.

Use the linear approximation

[{fleft( x right) approx Lleft( x right) }={ fleft( a right) + f^primeleft( a right)left( {x – a} right).}]

Take the derivative:

[{f^primeleft( x right) = left( {{x^2} + cos x} right)^prime }={ 2x – 2sin x.}]

Then

[{fleft( a right) = fleft( 0 right) }={ {0^2} + 2cos 0 }={ 2,}]

[{f^primeleft( a right) = f^primeleft( 0 right) }={ 2 cdot 0 – 2sin 0 }={ 0.}]

Substitute this into (Lleft( x right):)

[{Lleft( x right) = 2 + 0 cdot left( {x – 0} right) }={ 2.}]

Thus, the linear approximation of the given function is the horizontal line

[Lleft( x right) = 2.]

Solution.

Let (x = 179^circ), (a = 180^circ.) Hence, (Delta x = x – a = 179^circ – 180^circ = – 1) (= – {largefrac{pi }{{180}}normalsize}) radians. Calculate the value of the function and its derivative at the point (a:)

[
{fleft( {a} right) = sin 180^circ = 0,};;;kern-0.3pt
{f'left( x right) = {left( {sin x} right)^prime } = cos x,};;;kern-0.3pt
{f'left( {a} right) = cos 180^circ = – 1.}
]

Replacing the increment of the function by its differential, we obtain:

[
{fleft( x right) approx fleft( {a} right) + dy }
= {fleft( {a} right) + f'left( {a} right)Delta x,;;}Rightarrow
{sin 179^circ = 0 – 1 cdot left( { – frac{pi }{{180}}} right) }
= {frac{pi }{{180}} approx 0,0175.}
]

Solution.

We need to calculate (fleft( a right)) and (f^primeleft( a right)) where (a = 1.)

[{fleft( a right)} = fleft( 1 right) = ln 1 = {0},]

[f^primeleft( x right) = left( {ln x} right)^prime = frac{1}{x},]

[{f^primeleft( a right)} = f^primeleft( 1 right) = frac{1}{1} = {1}.]

Write the linear approximation function (Lleft( x right):)

Fasttasks 2 46 – The Troubleshooting Approximate Error

[{Lleft( x right) }={ fleft( a right) + f^primeleft( a right)left( {x – a} right) }={ 0 + 1left( {x – 1} right) }={ x – 1.}]

So the linearization of the natural logarithm at (x = 1) is

[{ln x approx x – 1.}]

Solution.

Consider the natural logarithmic function (y = ln x.) Given that

[ln {e^3} = 3ln e = 3,]

it is convenient to take the point ({a}) such that

[a = {e^3} approx 20,086.]

Calculate the derivative and its value at the point ({a}:)

[
{f'left( x right) = {left( {ln x} right)^prime } = frac{1}{x},;;}Rightarrow
{f'left( {a} right) approx frac{1}{{20,086}} approx 0,0498.}
]

Hence, the approximate value of (ln 20) is equal

[
{ln 20 approx ln {e^3} + 0,0498 cdot left( {20 – 20,086} right) }
= {3 – 0,0043 = 2,9957.}
]

Solution.

Let (fleft( x right) = {e^x}.) By setting (a = 0,) we have:

[
{fleft( {a} right) = {e^0} = 1,;;;}kern-0.3pt
{f'left( x right) = {left( {{e^x}} right)^prime } = {e^x},;;;}kern-0.3pt
{f'left( {a = 0} right) = {e^0} = 1.}
]

To calculate ({e^{0,1}}) we use the approximate formula

[fleft( x right) approx fleft( {a} right) + f'left( {a} right)left( {x – a} right).]

Then

[{e^{0,1}} approx 1 + 1 cdot left( {0,1 – 0} right) = 1,1.]

Solution.

Suppose that (fleft( x right) = arccos x) and (a = 0,5.) Replacing the increment of the function (Delta y) by its differential, we can compute the approximate value of (arccos 0,51:)

[
{fleft( x right) approx fleft( {a} right) + dy }
= {fleft( {a} right) + f'left( {a} right)left( {x – a} right),}
]

[
{f'left( x right) = {left( {arccos x} right)^prime } }
= { – frac{1}{{sqrt {1 – {x^2}} }},;;}Rightarrow
{f'left( {a = 0,5} right) }
= { – frac{1}{{sqrt {1 – 0,{5^2}} }} }
= { – frac{1}{{sqrt {0,75} }} approx – 1,1547,;;}Rightarrow
{arccos 0,51 }approx{ arccos 0,5 + left( { – 1,1547} right) cdot left( {0,51 – 0,5} right) }
= {frac{pi }{3} – 0,0115 } approx {1,035;text{radians}}approx {59,34^circ.}
]

Solution.

Let (fleft( x right) = arctan x), (a = 1.) Determine the value of the derivative at (a:)

[
{f'left( x right) = {left( {arctan x} right)^prime } = frac{1}{{1 + {x^2}}},;;}Rightarrow
{f'left( {a = 1} right) = frac{1}{{1 + {1^2}}} = frac{1}{2}.}
]

For the approximate calculation we use the formula

[fleft( x right) approx fleft( {a} right) + f'left( {a} right)left( {x – a} right).]

Consequently,

[
{arctan 0,95 approx arctan 1 + frac{1}{2} cdot left( {0,95 – 1} right) }
= {frac{pi }{4} + frac{1}{2} cdot left( { – 0,05} right) }
approx {0,7604;text{radians}} approx {43,57^circ.}
]

Solution.

We apply the linear approximation formula

[{fleft( x right) approx Lleft( x right) }={ fleft( a right) + f^primeleft( a right)left( {x – a} right),}]

where (a= 2.)

Take the derivative:

[{f^primeleft( x right) = left( {frac{3}{{{x^2}}}} right)^prime }={ left( {3{x^{ – 2}}} right)^prime }={ – 6{x^{ – 3}} }={ – frac{6}{{{x^3}}}.}]

The function and the derivative have the following values at (a = 2:)

[{fleft( a right) = fleft( 2 right) }={ frac{3}{{{2^2}}} }={ frac{3}{4},}]

[{f^primeleft( a right) = f^primeleft( 2 right) }={ – frac{6}{{{2^3}}} }={ – frac{3}{4}.}]

Plugging (a,) (fleft( a right),) and (f^primeleft( a right),) we obtain:

[{Lleft( x right) }={ frac{3}{4} + left( { – frac{3}{4}} right)left( {x – 2} right) }={ – frac{3}{4}x + frac{9}{4} }={ frac{3}{4}left( {3 – x} right).}]

Then the approximate value of (fleft( {1.99} right)) is equal to

[{fleft( {1.99} right) approx Lleft( {1.99} right) }={ frac{3}{4}left( {3 – 1.99} right) }={ 0.7575}]

Solution.

Choose the point (a = 1.) Then

[fleft( {a} right) = sqrt {{1^2} + 3 cdot 1} = 2.]

Find the value of the derivative of the given function at (a:)

[
{f'left( x right) = {left( {sqrt {{x^2} + 3x} } right)^prime } }
= {frac{1}{{2sqrt {{x^2} + 3x} }} cdot {left( {{x^2} + 3x} right)^prime } }
= {frac{{2x + 3}}{{2sqrt {{x^2} + 3x} }},;;}Rightarrow
{f'left( {a = 1} right)} = {frac{{2 cdot 1 + 3}}{{2sqrt {{1^2} + 3 cdot 1} }} }
= {frac{5}{4} = 1,25.}
]

Hence, the approximate value of the function at (x = 1,02) is equal to

[
{fleft( x right) approx fleft( {a} right) + f'left( {a} right)left( {x – a} right),;;}Rightarrow
{sqrt {1,02} approx 2 + 1,25 cdot left( {1,02 – 1} right)} = {2,025.}
]

Solution.

Let (a = 2). Consequently,

[fleft( {a = 2} right) = sqrt {5 cdot 2 – 1} = sqrt 9 = 3.]

The derivative at (a = 2) has the following value:

[
{f'left( x right) = {left( {sqrt {5x – 1} } right)^prime } }
= {frac{1}{{2sqrt {5x – 1} }} cdot {left( {5x – 1} right)^prime } }
= {frac{5}{{2sqrt {5x – 1} }},;;}Rightarrow
{f'left( {a = 2} right) }
= {frac{5}{{2sqrt {5 cdot 2 – 1} }} }
= {frac{5}{6} approx 0,833.}
]

Estimating the approximate value of the function at (x = 1,99), we have

[
{{fleft( x right) approx fleft( {a} right) + f'left( {a} right)left( {x – a} right),;}}Rightarrow
{{sqrt {1,99}} approx {3 + 0,833 cdot left( {1,99 – 2} right) }}
= {3 – 0,0083 approx 2,992.}
]

https://coolffil263.weebly.com/play-sin-city.html. Solution.

Let's linearize this function at (a = 3.) First, take the derivative using the quotient rule and the chain rule.

[{f^primeleft( x right) = left( {frac{{{x^2} + 1}}{{x – 2}}} right)^prime }={ frac{{2xleft( {x – 2} right) – left( {{x^2} + 1} right)}}{{{{left( {x – 2} right)}^2}}} }={ frac{{2{x^2} – 4x – {x^2} – 1}}{{{{left( {x – 2} right)}^2}}} }={ frac{{{x^2} – 4x – 1}}{{{{left( {x – 2} right)}^2}}}.}]

Calculate (fleft( a right)) and (f^primeleft( a right):)

[fleft( 3 right) = frac{{{3^2} + 1}}{{3 – 2}} = 10,]

Fasttasks 2 46 – The Troubleshooting Approximate The Number

[{f^primeleft( 3 right) = frac{{{3^2} – 4 cdot 3 – 1}}{{{{left( {3 – 2} right)}^2}}} }={ – 4.}]

Now we can write the linear approximation equation:

[{fleft( x right) approx Lleft( x right) }={ fleft( a right) + f^primeleft( a right)left( {x – a} right) }={ 10 – 4left( {x – 3} right) }={ – 4x + 22.}]

Thus, the linearization of the given function at (a = 3) is

[y = – 4x + 22.]

Example 25.

The function (fleft( x right) = – {x^2} + 100x) is linearized at a point (Pleft( {a,fleft( a right)} right).) The linear approximation line (Lleft( x right)) passes through (P) and intersects the (x-)axis at (x = 180.)

1. Find the coordinates (left( {a,fleft( a right)} right)) of the point (P.)
2. Write the linearization equation (Lleft( x right).)

Solution.

(1.) The linear approximation (Lleft( x right)) is given by the equation

[Lleft( x right) = fleft( a right) + f^primeleft( a right)left( {x – a} right).]

Take the derivative:

[{f^primeleft( x right) = left( { – {x^2} + 100x} right)^prime }={ – 2x + 100.}]

At the point (P,) the equation for (Lleft( x right)) becomes

[require{cancel}{Lleft( x right) }={ – {a^2} + 100a + left( { – 2a + 100} right)left( {x – a} right) }={ – {a^2} + cancel{100a} – 2ax + 100x + 2{a^2} – cancel{100a} }={ {a^2} + left( {100 – 2a} right)x.}]

We find the value of (a) from the condition (Lleft( x right) = 0) at (x = 180.) This yields:

[{{a^2} + left( {100 – 2a} right) cdot 180 = 0,;;} Rightarrow {{a^2} – 360a + 18000 = 0.}]

Solve the quadratic equation:

[D = {360^2} – 4 cdot 18000 = 57600.]

[{a = frac{{360 pm sqrt {57600} }}{2} }={ frac{{360 pm 240}}{2} }={ 60,300.}] Macos sierra 10 12 public beta 6 (16a304a) direct links.

We see that only one root (a = 60) belongs to the interval (left( {0,100} right),) so the point (P) has the coordinates:

[{a = 60,;}kern0pt{fleft( a right) = – {60^2} + 100 cdot 60 = 2400.}]

(2.) The equation of the linear approximation (Lleft( x right)) derived in Section (A) has the form

[Lleft( x right) = {a^2} + left( {100 – 2a} right)x.]

Substituting (a = 60,) we get

[{Lleft( x right) }={ {60^2} + left( {100 – 2 cdot 60} right)x }={ 3600 – 20x.}]

Fasttasks 2 46 – The Troubleshooting Approximate Square