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3 years ago

For what values of a and b is the line 3 x + y = b tangent to the curve y = a x^{3} when x = -2?

1 answer:

5 0

Hello,

The tangent point T is (-2,a(-2)^3) or (-2,b-3*(-2))
The slope of the curve in T is -3 (slope of the line)

y=ax^3==>y'=3ax²==>y'(-2)=3*a*4=-3 ==>4a=-1 ==>a=-1/4

T=(-2,-1/4*(-2)^3))=(-2,2)

b-3*(-2)=2==>b=2-6==>b=-4

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Solve: 2cos(x)-square root 3=0 for 0 less than x less than 2 pi

Leona [35]

Answer:

The general solution of $cos x = cos(\frac{\pi }{6})$ is

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The general solution values

$x = - \frac{\pi }{6} and x = \frac{\pi }{6}$

Step-by-step explanation:

Explanation:-

Given equation is

$2cosx-\sqrt{3} =0 for 0$

$2cosx =\sqrt{3}$

Dividing '2' on both sides, we get

$cos x =\frac{\sqrt{3} }{2}$

$cos x = cos(\frac{\pi }{6})$

General solution of cos θ = cos ∝ is θ = 2nπ±∝

Now The general solution of $cos x = cos(\frac{\pi }{6})$ is

x = 2nπ± $\frac{\pi }{6}$

put n=0

$x = - \frac{\pi }{6} and x = \frac{\pi }{6}$

Put n=1

$x = 2\pi +\frac{\pi }{6} = \frac{13\pi }{6}$

$x = 2\pi -\frac{\pi }{6} = \frac{11\pi }{6}$

put n=2

$x = 4\pi +\frac{\pi }{6} = \frac{25\pi }{6}$

$x = 4\pi -\frac{\pi }{6} = \frac{23\pi }{6}$

And so on

But given 0 < x< 2π

The general solution values

$x = - \frac{\pi }{6} and x = \frac{\pi }{6}$

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Step-by-step explanation:

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