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jenyasd209 [6]
3 years ago
14

Write the linearization of the function at the points indicated. (Enter your answer as an equation. Let x be the independent var

iable and y be the dependent variable.) x= \(\sqrt{?}25+x\) (0, 5) and (75, 10)
Mathematics
1 answer:
Natalija [7]3 years ago
5 0

Answer:

\displaystyle y=5+\frac{x}{10}

\displaystyle y=10+\frac{(x-75)}{20}

Step-by-step explanation:

<u>Linearization</u>

It consists of finding an approximately linear function that behaves as close as possible to the original function near a specific point.

Let y=f(x) a real function and (a,f(a)) the point near which we want to find a linear approximation of f. If f'(x) exists in x=a, then the equation for the linearization of f is

y=f(x)=f(a)+f'(a)(x-a)

Let's find the linearization for the function

y=\sqrt{25+x}

at (0,5) and (75,10)

Computing f'(x)

\displaystyle f'(x)=\frac{1}{2\sqrt{25+x}}

At x=0:

\displaystyle f'(0)=\frac{1}{2\sqrt{25+0}}=\frac{1}{10}

We find f(0)

f(0)=\sqrt{25+0}=5

Thus the linearization is

\displaystyle y=f(0)+f'(0)(x-0)=5+\frac{1}{10}x

\displaystyle y=5+\frac{x}{10}

Now at x=75:

\displaystyle f'(75)=\frac{1}{2\sqrt{25+75}}=\frac{1}{20}

We find f(75)

f(75)=\sqrt{25+75}=10

Thus the linearization is

\displaystyle y=f(75)+f'(75)(x-75)=10+\frac{1}{20}(x-75)

\displaystyle y=10+\frac{(x-75)}{20}

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Answer:

7)  C) PQR ≈ TSR

8) B)  3

9) D) 102 ft

10) D) 21

11) C) 45

Step-by-step explanation:

7) C) PQR ≈ TSR

8) 2 * 6 = 12

so 4x = 12

      x = 3

9)

24/20 = x/85

x = 24 * 85 / 20

x = 202 ft

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QT/35 = 24/40

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QT = 35 * 3 / 5

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