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Otrada [13]
4 years ago
11

A. Write the equation of the line that represents the linear approximation to the following function at the given point a.

Mathematics
1 answer:
anastassius [24]4 years ago
6 0

Answer:

a) f(x) = 4 - x²

The linear approximation of the function at a=1 is

L(x) = 5 - 2x at a = 1

b) The graph of the function and the linear approximation at that point is attached to this solution.

The curve represent the real function,

f(x) = 4 - x²

The straight line represents the linear approximation of the function at a=1.

L(x) = 5 - 2x

The curve and the function evidently cross paths at x=1 and understandably so.

c) Using the linear approximation obtained at a = 1.

f(1.1) = 2.8

Using the actual function, the actual value of f(1.1) = 2.79

d) Percent error = 0.358%

Step-by-step explanation:

f(x) = 4 - x²

a) The linear approximation of the function at the given point is given as

L(x) = f(a) + f'(a) [x - a]

f(x) = 4 - x²

a = 1

f(a) = 4 - 1² = 3

f'(x) = -2x

f'(a) = -2(1) = -2

L(x) = f(a) + f'(a) [x - a]

L(x) = 3 + (-2)(x - 1)

L(x) = 3 -2x + 2

L(x) = 5 - 2x

L(x) = -2x + 5

f(x) = 4 - x²

L(x) = 5 - 2x at a = 1

b) The graph of the function and the linear approximation at that point is attached to this solution.

The curve represent the real function,

f(x) = 4 - x²

The straight line represents the linear approximation of the function at a=1.

L(x) = 5 - 2x

The curve and the function evidently cross paths at x=1 and understandably so.

c) Use the linear approx. to estimate the given fxn value.

f(1.1)

L(x) = 5 - 2x

L(1.1) = 5 - 2(1.1) = 2.8

Using the function, the actual value of f(1.1) = 4 - 1.1² = 2.79

d) Compute the percent error in your approximation, 100*Iapprox-exactI/IexactI, where the exact value is given by a calculator

Percent error

= 100% × (|approx - exact|)/exact

Approximated value = 2.8

Exact value = 2.79

Percent error = 100% × (2.8-2.79)/2.79

Percent error = 0.358%

Hope this Helps!!!

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<h3>            x = -9,  y = -13 </h3><h3>    or    x = 13,   y = 9</h3><h3>    or    x = -13,  y = -9</h3><h3>    or     x = 9,   y = 13</h3>

Step-by-step explanation:

x^2+y^2=250\\\\x^2-2xy+y^2+2xy=250\\\\(x-y)^2=250-2xy\\\\(x-y)^2=250-2\cdot117\\\\ (x-y)^2=16\\\\x-y=4\qquad\qquad\vee\qquad \qquad  x-y=-4\\\\x=4+y \qquad\qquad \vee\qquad\qquad x=-4+y\\\\(y+4)y=117\qquad\vee\qquad\quad (y-4)y=117\\\\y^2+4y-117=0\qquad\vee\qquad y^2-4y-117=0\\\\y=\dfrac{-4\pm\sqrt{4^2-4(-117)}}{2\cdot1}\qquad\vee\qquad y=\dfrac{4\pm\sqrt{4^2-4(-117)}}{2\cdot1}\\\\y=\dfrac{-4\pm\sqrt{16+468}}{2}\qquad\ \ \vee\qquad y=\dfrac{4\pm\sqrt{16+468}}{2}

y_1=\dfrac{-4-22}{2}\ ,\quad y_2=\dfrac{-4+22}{2}\ ,\quad y_3=\dfrac{4-22}{2}\ ,\quad y_4=\dfrac{4+22}{2}\\\\y_1=-13\ ,\qquad y_2=9\ ,\qquad\quad\qquad\ y_3=-9\ ,\qquad y_4=13\\\\x_{1,2}=4+y_{1,2}\qquad\qquad\qquad\qquad\qquad x_{3,4}=-4+y_{3,4}\\\\x_1=-9\ ,\qquad x_2=13\ ,\qquad\quad\qquad x_3=-13\ ,\qquad x_4=9

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