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2 years ago

What is the inverse of the function f(x) = 2x + 1?

Mathematics

1 answer:

soldier1979 [14.2K]2 years ago

8 0

Answer: The inverse of the linear function f(x)=2x+1 is f^(-1) (x) = (1/2)x-1/2

Solution

f(x)=2x+1

y=f(x)

y=2x+1

Isolating x: Subtracting 1 both sides of the equation:

y-1=2x+1-1

y-1=2x

Multiplying both sides of the equation by 1/2:

(1/2)(y-1)=(1/2)2x

(1/2)y-1/2=x

x=(1/2)y-1/2

Changing "x" by "f^(-1) (x)" and "y" by "x":

f^(-1) (x) = (1/2)x-1/2

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natta225 [31]

Answer:

b. circle; $2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0$

Step-by-step explanation:

The given conic has equation;

$x^2-5x+y^2=3$

We complete the square to obtain;

$(x-\frac{5}{2})^2+(y-0)^2=\frac{37}{4}$

This is a circle with center;

$(\frac{5}{2},0)$

This implies that;

$x=\frac{5}{2},y=0$

When the circle is rotated through an angle of $\theta=\frac{\pi}{3}$ ,

The new center is obtained using;

$x'=x\cos(\theta)+y\sin(\theta)$ and $y'=-x\sin(\theta)+y\cos(\theta)$

We plug in the given angle with x and y values to get;

$x'=(\frac{5}{2})\cos(\frac{\pi}{3})+(0)\sin(\frac{\pi}{3})$ and $y'=--(\frac{5}{2})\sin(\frac{\pi}{3})+(0)\cos(\frac{\pi}{3})$

This gives us;

$x'=\frac{5}{4} ,y'=\frac{5\sqrt{3} }{4}$

The equation of the rotated circle is;

$(x'-\frac{5}{4})^2+(y'-\frac{5\sqrt{3} }{4})^2=\frac{37}{4}$

Expand;

$(x')^2+(y')^2-\frac{5\sqrt{3} }{2}y'-\frac{5}{2}x'+\frac{25}{4} =\frac{37}{4}$

Multiply through by 4; to get

$4(x')^2+4(y')^2-10\sqrt{3}y'-10x'+25 =37$

Write in general form;

$4(x')^2+4(y')^2-10x'-10\sqrt{3}y'-12 =0$

Divide through by 2.

$2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0$

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3 years ago

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Naddik [55]

Your answer:
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2 years ago

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erastova [34]

Answer for this question would be yellow (-2,-3)

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2 years ago

What is the sum of the first seven terms of the geometric series where a1 = 10 and r = −2?

lbvjy [14]

Answer:

Step-by-step explanation:

<h3>Given GP with:</h3>

a1 = 10
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<h3>Solution</h3>

<u>Formula of sum:</u>

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<u>Substitute values:</u>

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2 years ago

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Y_Kistochka [10]

Answer:

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