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geniusboy [140]
3 years ago
14

Find the linearization L(x) of the function at a. f(x) = x^4 + 2x^2, a = 1

Mathematics
1 answer:
gregori [183]3 years ago
8 0

The equation of the tangent line at x=1 can be written in point-slope form as

... L(x) = f'(1)(x -1) +f(1)

The derivative is ...

... f'(x) = 4x^3 +4x

so the slope of the tangent line is f'(1) = 4+4 = 8.

The value of the function at x=1 is

... f(1) = 1^4 +2·1^2 = 3

So, your linearization is ...

... L(x) = 8(x -1) +3

or

... L(x) = 8x -5

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Answer:

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Step-by-step explanation:

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4 years ago
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lawyer [7]

==> I can't be sure of the equation.  There's a blank space
        between '5t' and '8', where an operation must be.
       I think the operation is more likely ' + ' (addition).  Brainly usually
       does show the ' - ' if it's subtraction, but goes blank if it's addition.
       Tell you what I'll do:  Since you sound so desperate, and you're being
       so generous with your points,  I'll show you how to figure it out both ways.

----------------------------------------

If it's subtraction:                 5t - 8 = 43

Add  8  to each side:              5t      = 51

Divide each side by  5 :            t     = 51/5

                                                 t     = 10 and 1/5

--------------------------------------------

If it's addition:                         5t + 8  =  43

Subtract  8  from each side:     5t        =  35

Divide each side by  5 :             t        =   7 

5 0
3 years ago
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Choose one card value and its additive inverse. Choose from the list below to write a real-world story problem that would model
Pani-rosa [81]

Answer:

the answer is a

Step-by-step explanation:

above sea level is positive and below sea level is negative

hope this helps

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3 years ago
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3 years ago
Find the equation of the tangent to the circle x2 + y2 = 109 at point (-10,3)
konstantin123 [22]

Answer:

10x - 3y = -109 .

Step-by-step explanation:

x^2 + y^2 = 109

Implicit Differentiation:

2x + 2y.y' = 0

y' = -2x/2y = -x/y

So at the point (-10,3) the gradient of the tangent is -(-10)/3 = 10/3.

Equation of the tangent:

y - y1 = m(x - x1)

y - 3 = 10/3(x + 10)

y - 3 = 10/3 x + 100/3

3y - 9 = 10x + 100

-109 + 3y = 10x

10x - 3y = -109  

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3 years ago
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