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## #f# is a function on #RR# where #f(x)=|x^2-2|#, How do you find #f^-1(x)#?

Parabola
Featured 3 months ago

#### Explanation:

Let's turn $f \left(x\right) = \left\mid {x}^{2} - 2 \right\mid$ into a piece-wise function.

Now, we see that $\left\mid a \right\mid = a$ when $a \ge 0$
Similarly, we see that $\left\mid a \right\mid = - a$ when $a < 0$

Therefore, we have:
$f \left(x\right) = {x}^{2} - 2$ when ${x}^{2} - 2 \ge 0$

$f \left(x\right) = - \left({x}^{2} - 2\right)$ when ${x}^{2} - 2 < 0$

Let's graph the two parabolas. When we do, we get:

Now, the blue parabola applies when ${x}^{2} - 2 \ge 0$, and green parabola applies when ${x}^{2} - 2 < 0$.

Therefore, we have(focus on the red graph):

Now, to find its inverse, we have to reflect the red graph over the line $y = x$.

Let's try this mathematically.

Our piece-wise function was:
$f \left(x\right) = {x}^{2} - 2$ when ${x}^{2} - 2 \ge 0$

$f \left(x\right) = - \left({x}^{2} - 2\right)$ when ${x}^{2} - 2 < 0$

Let's find the inverse function for each situation.

Now, remember that $f \left({f}^{-} 1 \left(x\right)\right) = x$ and ${f}^{-} 1 \left(f \left(x\right)\right) = x$

Therefore, we can now find the inverse functions.

$f \left(x\right) = {x}^{2} - 2 \implies x = {\left({f}^{-} 1 \left(x\right)\right)}^{2} - 2$

=>$x + 2 = {\left({f}^{-} 1 \left(x\right)\right)}^{2}$

=>$\pm \sqrt{x + 2} = {f}^{-} 1 \left(x\right)$ when ${x}^{2} - 2 \ge 0$

Similarly,

$f \left(x\right) = - {x}^{2} + 2 \implies x = - {\left({f}^{-} 1 \left(x\right)\right)}^{2} + 2$

=>$x - 2 = - {\left({f}^{-} 1 \left(x\right)\right)}^{2}$

=>$2 - x = {\left({f}^{-} 1 \left(x\right)\right)}^{2}$

=>$\pm \sqrt{2 - x} = {f}^{-} 1 \left(x\right)$ when ${x}^{2} - 2 < 0$

We now graph these two sideways parabolas:

When $y \ge \sqrt{2}$ or $y \le - \sqrt{2}$ , then we apply our first function.

When $- \sqrt{2} < y < \sqrt{2}$ , then we apply our second function.

We therefore have:

Just note that $\pm \sqrt{x + 2}$ and other relations like this one is not a function but a relation because there are more than one output for a given input.

## A(2,8), B(6,4) and C(-6,y) are collinear points find y?

Ratnaker Mehta
Featured 2 months ago

$16$.

#### Explanation:

Prerequisite :

$\text{The points "(x_1,y_1),(x_2,y_2) and (x_3,y_3)" are collinear}$

$\Leftrightarrow | \left({x}_{1} , {y}_{1} , 1\right) , \left({x}_{2} , {y}_{2} , 1\right) , \left({x}_{3} , {y}_{3} , 1\right) | = 0$.

Therefore, in our Problem, $| \left(2 , 8 , 1\right) , \left(6 , 4 , 1\right) , \left(- 6 , y , 1\right) | = 0$,

$\Rightarrow 2 \left(4 - y\right) - 8 \left\{6 - \left(- 6\right)\right\} + 1 \left\{6 y - \left(- 24\right)\right\} = 0$,

$\Rightarrow 8 - 2 y - 96 + 6 y + 24 = 0$,

$\Rightarrow 4 y = 64$,

$\Rightarrow y = 16 ,$ as Respected Lorenzo D. has already derived!.

## When the equation #y=5x+p# is a constant, is graphed in the #xy#-plane, the line passes through the point (-2,1). what is the value of p?

VNVDVI
Featured 2 months ago

$p = 11$

#### Explanation:

Our line is in the form of $y = m x + b$, where $m$ is the slope and $b$ is the $y$-coordinate of the $y$-intercept, $\left(0 , b\right)$ .

Here, we can see $m = 5$ and $b = p$.

Recall the formula for the slope:

$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

Where $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ are two points through which the line with this slope passes.

$m = 5$:

$5 = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

We are given a point through which the line passes, $\left(- 2 , 1\right)$, so $\left({x}_{1} , {y}_{1}\right) = \left(- 2 , 1\right)$

Since $b = p$, we know our $y$-intercept for this line is $\left(0 , p\right)$. The y-intercept is certainly a point through which the line passes. So, $\left({x}_{2} , {y}_{2}\right) = \left(0 , p\right)$

Let's rewrite our slope equation with all of this information:

$5 = \frac{p - 1}{0 - \left(- 2\right)}$

We now have an equation with one unknown variable, $p ,$ for which we can solve:

$5 = \frac{p - 1}{2}$

$5 \left(2\right) = \left(p - 1\right)$

$10 = p - 1$

$p = 11$

## Factoring Trinomials 5x^2+3x+4?

Nimo N.
Featured 2 months ago

As was stated in Nghi N's answer, the polynomial can not be factored... using real numbers.
However, if you are in a more advanced algebra course, and you have met complex numbers, see below.

#### Explanation:

Problem:
Factor: # color(blue)( 5x^2 + 3x + 4 #

The discriminant is negative
$D = \left({b}^{2} - 4 a c\right) = - 71$, so the quadratic formula will produce a square root of a negative number, which isn't a real number.

However, the quadratic formula gives roots, which can then be written in the form of factors.

• If r is a root for a polynomial expression, then (x - r) is a factor. This is true, whether or not the roots are real.

# color(blue)( x = ( (-b) +- sqrt(b^2 - 4ac) )/(2a)#
The value of the expression inside the square root is already known, # color(brown)( D = -71 #. Fill-in the rest of the numbers to get the two roots.

The roots:
$x = \frac{\left(- 3\right) \pm \sqrt{- 71}}{2 \cdot 5}$
# color(brown)( x = ( -3 +- sqrt(-71) )/(10) #

The two factors are:
$\left(x - \frac{- 3 + \sqrt{- 71}}{10}\right)$
# color(brown)( ( x + ( 3 - sqrt(-71) )/(10) ) #,
and
$x - \frac{- 3 - \sqrt{- 71}}{10}$
# color(brown)( ( x + ( 3 + sqrt(-71) )/(10) ) #

Here's where the complex numbers appear:
# sqrt( -71 ) = sqrt( 71 * (-1) ) = sqrt( 71 )* sqrt(-1) ) #
# sqrt( -71 ) = sqrt( 71 ) * i ) #, which is usually written as:
# color(brown)( sqrt( -71 ) = i * sqrt( 71 ) #

Substitute the value above to obtain the two "complex factors" of the polynomial:
# color(brown)( ( x + ( 3 - i *sqrt(71) )/(10) ) #, and

# color(brown)( ( x + ( 3 + i * sqrt(71) )/(10) ) #

## Solve the polynomial inequality and express in interval notation? x^2-2x-15 < 0 I don't understand why the answer (-3,5) does not include negative infinity or infinity in the answer

Tazwar Sikder
Featured 2 months ago

$- 3 < x < 5$

#### Explanation:

We have: ${x}^{2} - 2 x - 15 < 0$

$R i g h t a r r o w {x}^{2} + 3 x - 5 x - 15 < 0$

$R i g h t a r r o w x \left(x + 3\right) - 5 \left(x + 3\right) < 0$

$R i g h t a r r o w \left(x + 3\right) \left(x - 5\right) < 0$

Now, for a product to be less than zero, either one factor must be less than zero and the other greater than zero, i.e. either one factor is positive and one is negative, or vice versa:

$R i g h t a r r o w x + 3 < 0$ and $x - 5 > 0 R i g h t a r r o w x < - 3$ and $x > 5$

or

$R i g h t a r r o w x - 5 < 0$ and $x + 3 > 0 R i g h t a r r o w x < 5$ and $x > - 3$

The first case is not possible ($x$ cannot be greater than $5$ and less than $- 3$).

Therefore, the solution to the inequality is $- 3 < x < 5$.

"Negative" and "positive" infinity aren't included in the answer because $x$ lies between, and only between, $- 3$ and $5$.

## How do you factor completely #x^3 + y^3#?

George C.
Featured 1 month ago

${x}^{3} + {y}^{3} = \left(x + y\right) \left({x}^{2} - x y + {y}^{2}\right)$

$\textcolor{w h i t e}{{x}^{3} + {y}^{3}} = \left(x + y\right) \left(x - \left(\frac{1}{2} + \frac{\sqrt{3}}{2} i\right) y\right) \left(x - \left(\frac{1}{2} - \frac{\sqrt{3}}{2} i\right) y\right)$

#### Explanation:

Given:

${x}^{3} + {y}^{3}$

Note that if $y = - x$ then this is zero. So we can deduce that $\left(x + y\right)$ is a factor and separate it out:

${x}^{3} + {y}^{3} = \left(x + y\right) \left({x}^{2} - x y + {y}^{2}\right)$

We can calculate the discriminant for the remaining homogeneous quadratic in $x$ and $y$ just like we would for a quadratic in a single variable:

${x}^{2} - x y + {y}^{2}$

is in standard form:

$a {x}^{2} + b x y + c {y}^{2}$

with $a = 1$, $b = - 1$ and $c = 1$

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\textcolor{b l u e}{1}}^{2} - 4 \left(\textcolor{b l u e}{1}\right) \left(\textcolor{b l u e}{1}\right) = - 3$

Since $\Delta < 0$, this quadratic has no linear factors with real coefficients.

We can factor it with complex coefficients by completing the square and using ${i}^{2} = - 1$ as follows:

${x}^{2} - x y + {y}^{2} = {\left(x - \frac{1}{2} y\right)}^{2} + \frac{3}{4} {y}^{2}$

$\textcolor{w h i t e}{{x}^{2} - x y + {y}^{2}} = {\left(x - \frac{1}{2} y\right)}^{2} + {\left(\frac{\sqrt{3}}{2} y\right)}^{2}$

$\textcolor{w h i t e}{{x}^{2} - x y + {y}^{2}} = {\left(x - \frac{1}{2} y\right)}^{2} - {\left(\frac{\sqrt{3}}{2} y i\right)}^{2}$

$\textcolor{w h i t e}{{x}^{2} - x y + {y}^{2}} = \left(\left(x - \frac{1}{2} y\right) - \frac{\sqrt{3}}{2} i y\right) \left(\left(x - \frac{1}{2} y\right) + \frac{\sqrt{3}}{2} i y\right)$

$\textcolor{w h i t e}{{x}^{2} - x y + {y}^{2}} = \left(x - \left(\frac{1}{2} + \frac{\sqrt{3}}{2} i\right) y\right) \left(x - \left(\frac{1}{2} - \frac{\sqrt{3}}{2} i\right) y\right)$

So:

${x}^{3} + {y}^{3} = \left(x + y\right) \left({x}^{2} - x y + {y}^{2}\right)$

$\textcolor{w h i t e}{{x}^{3} + {y}^{3}} = \left(x + y\right) \left(x - \left(\frac{1}{2} + \frac{\sqrt{3}}{2} i\right) y\right) \left(x - \left(\frac{1}{2} - \frac{\sqrt{3}}{2} i\right) y\right)$

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