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prisoha [69]
3 years ago
8

Find the derivative of f(x) = 6/x at x = -2. select one: a. 6 b. 3/2 c. -3/2 d. -6

Mathematics
1 answer:
Ivenika [448]3 years ago
5 0
F(X) = 6/X

F(X) = 6 • 1/X

F(X) = 6 • x^-1

F(X) = 6x^-1

F'(X) = 6 • d(x^-1)/dx

F'(X) = 6 • -1x^-1-1

F'(X) = 6 • -1x^-2

F'(X) = -6x^-2

F'(X) = -6/x^2
F'(-2) = -6/(-2)^2
F'(-2) = -6/4
F'(-2) = -3/2

The solution would be C. -3/2.
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<h3>Answer:  A.  18*sqrt(3)</h3>

=============================================

Explanation:

We'll need the tangent rule

tan(angle) = opposite/adjacent

tan(R) = TH/HR

tan(30) = TH/54

sqrt(3)/3 = TH/54 ... use the unit circle

54*sqrt(3)/3 = TH .... multiply both sides by 54

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TH = 18*sqrt(3) which points to <u>choice A</u> as the final answer

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

An alternative method:

Triangle THR is a 30-60-90 triangle.

Let x be the measure of side TH. This side is opposite the smallest angle R = 30, so we consider this the short leg.

The hypotenuse is twice as long as x, so TR = 2x. This only applies to 30-60-90 triangles.

Now use the pythagorean theorem

a^2 + b^2 = c^2

(TH)^2 + (HR)^2 = (TR)^2

(x)^2 + (54)^2 = (2x)^2

x^2 + 2916 = 4x^2

2916 = 4x^2 - x^2

3x^2 = 2916

x^2 = 2916/3

x^2 = 972

x = sqrt(972)

x = sqrt(324*3)

x = sqrt(324)*sqrt(3)

x = 18*sqrt(3) which is the length of TH.

A slightly similar idea is to use the fact that if y is the long leg and x is the short leg, then y = x*sqrt(3). Plug in y = 54 and isolate x and you should get x = 18*sqrt(3). Again, this trick only works for 30-60-90 triangles.

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3 years ago
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