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4 years ago

Let f(x) = √x-5 Find f^ -1.

Mathematics

1 answer:

iris [78.8K]4 years ago

5 0

Answer:

$f^{-1}(x)= x^2 + 5 \,,\,\,x\geq 0$

Step-by-step explanation:

Use the method of swapping x and y and isolating y.

$f(x) = y = \sqrt{x-5}\,,\,\,\,x\geq 5,y\geq0\\\\f^{-1}:x=\sqrt{y-5}\,,\,\,\,y\geq 5, x\geq 0\\x^2 = y-5\\y = x^2 + 5 = f^{-1}(x)\\$

Note that domain/range will be swapped as well.

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Find the sum of the roots of the quadratic x^2 + 7x - 13 = 0

Hoochie [10]

Answer:

-7

Step-by-step explanation:

You use the quadratic formula of $ax^{2} +bx+c = 0$ to get:

a = 1

b = 7

c = -13

then you plug that into x= (-b ± √b²-4ac)/(2a) :

x= ( -7 ± √7²- 4(1)(-13) )/(2(1))

$x=-\frac{7}{2} + \frac{\sqrt{101}}{2} \\\\x=-\frac{7}{2} - \frac{\sqrt{101}}{2}$

The sum of these two x values turns out to be:

$x=-\frac{7}{2} + \frac{\sqrt{101}}{2}+( -\frac{7}{2} - \frac{\sqrt{101}}{2} )= -\frac{14}{2} = 7$

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3 years ago

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grigory [225]

Answer:

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Step-by-step explanation:

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3 years ago

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strojnjashka [21]

We don't really have to put this in slope intercept form; it would clearly end up as

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so we know the slope is -4.

Answer: C) -4

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3 years ago

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castortr0y [4]

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The directrix of a parabola is x = 4. Its focus is (2,6) .

Sati [7]

Answer:

Step-by-step explanation:

If you plot the directrix and the focus, you can see that the focus is to the left of the directrix. Since a parabola ALWAYS wraps itself around the focus, our parabola opens sideways, to the left to be specific. The formula for the parabola that opens to the left is

$-(y-k)^2=4p(x-h)$

We will solve this for x at the end. The negative is out front because it opens to the left. If it opened to the right, it would be positive.

The vertex of a parabola is exactly halfway between the focus and the directrix, so our vertex coordinates h and k are (3, 6). P is defined as the distance between the vertex and the directrix, or the vertex and the focus. Since the vertex is directly between both the directrix and the focus, each distance is the same. P = 1. Filling in what we have now:

$-(y-6)^2=4(1)(x-3)$ which simplifies to

$-(y-6)^2=4(x-3)$

Now we will solve it for x.

$-(y-6)^2=4x-12$ and

$-(y-6)^2+12=4x$ so

$-\frac{1}{4}(y-6)^2+12=x$

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4 years ago