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Leno4ka [110]
2 years ago
13

Given f"(x) = 2, f'(1)=4, and f(2)=-2, find f(x).

Mathematics
1 answer:
Anna [14]2 years ago
6 0

Answer:

Step-by-step explanation:

f"(x)=2

integrating

f'(x)=2x+c

f'(1)=2+c=4

c=4-2=2

f'(x)=2x+2

integrating

f(x)=2x^2/2+2x+a

f(x)=x^2+2x+a

f(2)=-2

(2)^2+2(2)+a=-2

4+4+a=-2

a=-2-8=-10

f(x)=x^2+2x-10

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RSB [31]

Step-by-step explanation:

a) f(1) = -1 → x = -1

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2 years ago
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WARRIOR [948]
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Read 2 more answers
3cis(5pi/4) in polar form
madreJ [45]
Rectangular form:
z = -2.1213203-2.1213203i

Angle notation (phasor):
z = 3 ∠ -135°

Polar form:
z = 3 × (cos (-135°) + i sin (-135°))

Exponential form:
z = 3 × ei (-0.75) = 3 × ei (-3π/4)

Polar coordinates:
r = |z| = 3 ... magnitude (modulus, absolute value)
θ = arg z = -2.3561945 rad = -135° = -0.75π = -3π/4 rad ... angle (argument or phase)

Cartesian coordinates:
Cartesian form of imaginary number: z = -2.1213203-2.1213203i
Real part: x = Re z = -2.121
Imaginary part: y = Im z = -2.12132034
8 0
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