Find the inverse of the function y = 2x2 + 2

2 answers:

8 0

We have that
y = 2x² + 2

step 1

exchange the value of x for y and the value of y for x
y = 2x² + 2------> x=2y²+2

step 2
clear y variable
x=2y²+2----> 2y²=x-2----> y²=[(x-2)/2]-----> y=(+/-)√[(x-2)/2]

step 3
the inverse is
f(x)-1=(+/-)√[(x-2)/2]

VMariaS [17]3 years ago

8 0

Answer:

y=+- $\sqrt{\frac{1}{2} x-1}$

Step-by-step explanation:

You might be interested in

Verify identity: (sec(x)-csc(x))/(sec(x)+csc(x))=(tan(x)-1)/(tan(x)+1)

Nikitich [7]

So hmmm let's do the left-hand-side first

$\bf \cfrac{sec(x)-csc(x)}{sec(x)+csc(x)}\implies \cfrac{\frac{1}{cos(x)}-\frac{1}{sin(x)}}{\frac{1}{cos(x)}+\frac{1}{sin(x)}}\implies 
\cfrac{\frac{sin(x)-cos(x)}{cos(x)sin(x)}}{\frac{sin(x)+cos(x)}{cos(x)sin(x)}}
\\\\\\
\cfrac{sin(x)-cos(x)}{cos(x)sin(x)}\cdot \cfrac{cos(x)sin(x)}{sin(x)+cos(x)}\implies \boxed{\cfrac{sin(x)-cos(x)}{sin(x)+cos(x)}}$

now, let's do the right-hand-side then

$\bf \cfrac{tan(x)-1}{tan(x)+1}\implies \cfrac{\frac{sin(x)}{cos(x)}-1}{\frac{sin(x)}{cos(x)}+1}\implies \cfrac{\frac{sin(x)-cos(x)}{cos(x)}}{\frac{sin(x)+cos(x)}{cos(x)}}
\\\\\\
\cfrac{sin(x)-cos(x)}{cos(x)}\cdot \cfrac{cos(x)}{sin(x)+cos(x)}\implies \boxed{\cfrac{sin(x)-cos(x)}{sin(x)+cos(x)}}$

7 0

3 years ago

PLZZZZZZ NEED HELP!!!!!!!!!!

egoroff_w [7]

The first one is the second one. t(n)=20n+480

Question 4: is the first one t(n)=8n+15

4 0

3 years ago

Explain how to check multiplication using addition and division ñ. Include an example in your explanation.

ratelena [41]

Addition/Subtraction and. Multiplication/ Division ... So checking one operation with its opposite is a good way to help verify your answer.

5 0

3 years ago