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3 years ago

14

Email used a total of 7 1half

1 answer:

Gala2k [10]3 years ago

7 0

13 - 5 5/6
= 12 6/6 - 5 5/6
= 7 1/6

hope it helps

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Given f(x) = 4x^4 find f^-1(x) Then state whether f^-1(x) is a function.

sdas [7]

Answer:

$f^{-1}(x)=\pm\sqrt[4]{\dfrac{x}{4}}$
$f^{-1}(x) \quad\text{is not a function}$

Step-by-step explanation:

To find the inverse function, solve for y:

$x=f(y)\\\\x=4y^4\\\\\dfrac{x}{4}=y^4\\\\\pm\sqrt[4]{\dfrac{x}{4}}=y\\\\f^{-1}(x)=\pm\sqrt[4]{\dfrac{x}{4}}$

f(x) is an even function, so f(-x) = f(x). Then the inverse relation is double-valued: for any given y, there can be either of two x-values that will give that result.

___

A function is single-valued. That means any given domain value maps to exactly one range value. The test of this is the "vertical line test." If a vertical line intersects the graph in more than one point, then that x-value maps to more than one y-value.

The horizontal line test is similar. It is used to determine whether a function has an inverse function. If a horizontal line intersects the graph in more than one place, the inverse relation is not a function.

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Since the inverse relation for the given f(x) maps every x to two y-values, it is not a function. You can also tell this by the fact that f(x) is an even function, so does not pass the horizontal line test. When f(x) doesn't pass the horizontal line test, f^-1(x) cannot pass the vertical line test.

_____

The attached graph shows the inverse relation (called f₁(x)). It also shows a vertical line intersecting that graph in more than one place.

5 0

2 years ago

The function Q(x)=2x^2-kx+18. For what value of k does Q(x) have one distinct real solution? (NEEDS TO BE DONE WITHIN 2 HOURS!)

Aliun [14]

Answer:

k = 12

Step-by-step explanation:

Given:

The equation $Q(x)=2x^2-kx+18$

To find:

Value of $k = ?$ for which the given equation has one distinct real solution.

Solution:

The given equation is a quadratic equation.

There are always two solutions of a quadratic equation.

For the equation: $ax^{2} +bx+c=0$ to have one distinct solution:

$b^2 - 4ac = 0$

Here,

a = 2,

b = -k and

c = 18

Putting the values, we get:

$(-k)^2 - 4\times 2\times 18 = 0\\\Rightarrow k^2 = 18\times 8\\\Rightarrow k^2 =144\\\Rightarrow k = 12$

The equation becomes:

$Q(x)=2x^2-12x+18$

And the one root is:

$2(x^2-6x+9 ) = 0\\\Rightarrow 2(x-3)^2=0\\\Rightarrow x = 3$

4 0

3 years ago

Substitution:<br><br> y=6x+11<br><br>y-4x=15

german

Answer:

x=2

Step-by-step explanation:

6x+11-4x=15

2x+11=15

2x=4

x=2

4 0

3 years ago

Please find the question

melisa1 [442]

Answer:

By distance formula ,

AB² = (5+1)² + (3-6)²

AB² = 36 + 9

AB = 6.71 units

6 0

3 years ago

Determine whether the given function is a solution to the given differential equation.

lesantik [10]

Given :

A function , x = 2cos t -3sin t .....equation 1.

A differential equation , x'' + x = 0 .....equation 2.

To Find :

Whether the given function is a solution to the given differential equation.

Solution :

First derivative of x :

$x'=\dfrac{d(2cos t - 3sin t)}{dt}\\\\x'=\dfrac{d(2cost)}{dt}-\dfrac{(3sint)}{dt}\\\\x'=-2sint-3cost$

Now , second derivative :

$x''=\dfrac{d(-2sint-3cost)}{dt}\\\\x''=-\dfrac{d(2sint)}{dt}-\dfrac{d(3cost)}{dt}\\\\x''=-2cost+3sint$

( Note : derivative of sin t is cos t and cos t is -sin t )

Putting value of x'' and x in equation 2 , we get :

=(-2cos t + 3sin t ) + ( 2cos t -3sin t )

= 0

So , x'' and x satisfy equation 2.

Therefore , x function is a solution of given differential equation .

Hence , this is the required solution .

8 0

3 years ago