Mathematics: For the function f(x) -4 sqrtx - 1, find the inverse function.

For the function f(x) -4 sqrtx - 1, find the inverse

Ответ:

lopezsharon333

12.03.2022

Mathematics

$f^-^1(x)=\frac{(x+1)^2}{16}$

Step-by-step explanation:

We have the function: $f(x)=-4\sqrt{x} -1$ and we have to find the inverse.

Observation: y=f(x) , then $y=-4\sqrt{x} -1$

You can obtain the inverse of a function by switching the x and y values. This means:

$y=-4\sqrt{x} -1\\x=-4\sqrt{y'} -1$

We're going to call f^-^1(x)=y' . Now we have to clear "y'".

First we have to add (1) in both sides of the equation:

$x=-4\sqrt{y'} -1\\x+1=-4\sqrt{y'}-1+1\\x+1=-4\sqrt{y'}$

Now divide in (-4) both sides.

$x+1=-4\sqrt{y'}\\\frac{(x+1)}{(-4)} =\frac{(-4)\sqrt{y'}}{(-4)} \\\\\frac{(x+1)}{(-4)} =\sqrt{y'}$

And for our last step we have to square both sides:

$\frac{(x+1)}{(-4)} =\sqrt{y'}\\\\(\frac{(x+1)}{(-4)})^2=(\sqrt{y'})^2\\\frac{(x+1)^2}{(-4)^2}=y'\\\\\frac{(x+1)^2}{16}=y'$

Then the third option is the correct: $f^-^1(x)=\frac{(x+1)^2}{16}$

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Ответ:

gwendallinesikes

06.09.2022

Mathematics

3x2y + 3xy (cannot be added)

3x2y + (–12x2y) = -9x2y

3x2y + 2x2y2 (cannot be added)

3x2y + 7xy2 (cannot be added)

3x2y + (–10x2) (cannot be added)

3x2y + 4x2y = 7x2y

3x2y + 3x3 (cannot be added)

Step-by-step explanation:

Adding and subtracting a monomial requires having the same variables. No matter how big or small their coefficient is, if their variables do not match, they cannot be added or subtracted. The crucial part in adding or subtracting monomials is their sign. If the signs are the same, retain the sign. If the signs are different, subtract and keep the sign of the larger number.

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For the function f(x) -4 sqrtx - 1, find the inverse function.

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