2 years ago

Help!!! I’m bad at Math!

Mathematics

2 answers:

ollegr [7]2 years ago

8 0

3) -0.84, -0.32, 0.32, 0.84
4) -4/9, -1/3, 2/9, 1/3

e-lub [12.9K]2 years ago

6 0

-0.84 , -0.32, 0.32, 0.84

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Help me please!

telo118 [61]

Answer:

$f^{-1}(6) = 50$

Step-by-step explanation:

Given

$f(x) = \sqrt{2x} - 4$

Required

Find $f^{-1}(6)$

First, we calculate the inverse function

$f(x) = \sqrt{2x} - 4$

Express f(x) as y

$y = \sqrt{2x} - 4$

Swap the positions of x and y

$x = \sqrt{2y} - 4$

Solve for y: Add 4 to both sides

$4 + x = \sqrt{2y} - 4+4$

$4 + x = \sqrt{2y}$

Square both sides

$(4 + x)^2 = 2y$

Divide both sides by 2

$y = \frac{(4 + x)^2}{2}$

Express y as an inverse function

$f^{-1}(x) = \frac{(4 + x)^2}{2}$

Next, solve for: $f^{-1}(6)$

Substitute 6 for x

$f^{-1}(6) = \frac{(4 + 6)^2}{2}$

$f^{-1}(6) = \frac{(10)^2}{2}$

$f^{-1}(6) = \frac{100}{2}$

$f^{-1}(6) = 50$

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Step-by-step explanation:

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