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skelet666 [1.2K]
4 years ago
5

This work problem to solving Equations

Mathematics
1 answer:
Angelina_Jolie [31]4 years ago
8 0

Answer:

question por favor. ............ ....

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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.

__

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
3 years ago
Ethan ran 11 miles in 2 hours What is the constent of proporttional of miles to a hour
Dafna1 [17]

Answer:

5.5 miles

Step-by-step explanation:

If he ran 11 miles in 2 hours then divide both of them by 2, so 1 hour is equal to 11 / 2 which is 5.5 miles

5 0
2 years ago
Subtract 8-4 1/2. A. 3 B. 4 C. 4 1/2 D. 3 1/2
kicyunya [14]
This was the answer I came up with - 3.5
3 0
3 years ago
Find the average value of the function over the given interval and all values of x in the interval for which the function equals
MrRissso [65]

The average value of f(x)=4x^3-3x^2 over [-1, 3] is

\displaystyle\frac1{3-(-1)}\int_{-1}^3(4x^3-3x^2)\,\mathrm dx=\frac14\left(x^4-x^3\right)\bigg|_{-1}^3=\frac{(3^4-3^3)-(1+1)}4=\boxed{13}

8 0
3 years ago
How do I solve #9 & #10
olasank [31]
Do you mean the last two? the 9 and 10 are cut off
6 0
3 years ago
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