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3 years ago

Find the inverse function of (Show work) f(x)=x^2-4

Mathematics

1 answer:

Olegator [25]3 years ago

4 0

Answer:

The answer to your question is $f(x) = \sqrt{x+ 4}$

Step-by-step explanation:

f(x) = x² - 4

Process

1.- Change f(x) for y

y = x² - 4

2.- Change "x" for "y" and "y" for "x".

x = y² - 4

3.- Make "y" the object of the equation

y² = x + 4

$y = \sqrt{x+ 4}$

4.- Change "y" for f(x)

$f(x) = \sqrt{x+ 4}$

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Answer: 2x^3

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2 years ago

Jasmine launches a model rocket from a cliff. The rocket goes up 16.8 feet above the cliff, then it drops 28.1 feet into the oce

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Answer:

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Step-by-step explanation

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2 years ago

Let T:ℝ2→ℝ2 be the linear transformation that first rotates points clockwise through 45∘ (????/4 radians) and then reflects poin

Alisiya [41]

Answer:

$T = \left[\begin{array}{ccc}-\frac{1}{\sqrt{2} } &\frac{1}{\sqrt{2} }\\\frac{1}{\sqrt{2} }&\frac{1}{\sqrt{2} }\end{array}\right]$

Step-by-step explanation:

Let General Transformation matrix be denoted as T

Step 1: Clockwise rotation of 45 degrees

General counterclockwise rotation matrix in 2-dimension is given as

$R(\theta)=\left[\begin{array}{ccc}cos\theta & - sin\theta\\sin\theta&cos\theta\\\end{array}\right]$

For clockwise rotation we need to insert θ as negative in the above matrix. Therefore, the resulting matrix is

$R(-\theta)=\left[\begin{array}{ccc}cos\theta & sin\theta\\-sin\theta&cos\theta\\\end{array}\right]$

as sin(-θ) = -sin (θ) and cos(-θ) = cos (θ)

For 45 degrees

$sin(45) = \frac{1}{\sqrt{2} }$ and $cos(45) = \frac{1}{\sqrt{2} }$

$R(-45)=\left[\begin{array}{ccc}\frac{1}{\sqrt{2} } & \frac{1}{\sqrt{2} }\\-\frac{1}{\sqrt{2} }&\frac{1}{\sqrt{2} }\\\end{array}\right]$

Step 2: Reflection through line y = x

This type of reflection maps (x,y)→(y,x)

Therefore the general matrix is

$R(x,y)=\left[\begin{array}{ccc}0&1\\1&0\end{array}\right]$

Step 3: General Transformation Matrix

T = R(x,y) R(-θ)

$T=\left[\begin{array}{ccc}0&1\\1&0\end{array}\right] \left[\begin{array}{ccc}\frac{1}{\sqrt{2} } & \frac{1}{\sqrt{2} }\\-\frac{1}{\sqrt{2} }&\frac{1}{\sqrt{2} }\\\end{array}\right]$

$T = \left[\begin{array}{ccc}-\frac{1}{\sqrt{2} } &\frac{1}{\sqrt{2} }\\\frac{1}{\sqrt{2} }&\frac{1}{\sqrt{2} }\end{array}\right]$

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