No, they forgot to switch variable labels after solving for the independent variable...
y=-8x+4
y-4=-8x
(y-4)/-8=x
Now that you have solved for the independent variable x, you switch the variable labels...
y=(x-4)/-8
f^-1(x)=(x-4)/-8 which should really be rewritten as:
f^-1(x)=(4-x)/8 :P
Answer:
I think D
Step-by-step explanation:
Answer:
option (a) f(x)= 1/x+2
Step-by-step explanation:
(a) f(x) = 1/ x+2
To find the restriction for domain , we set the denominator =0 and solve for x
x+2 =0, so x=-2
When x=-2 then denominator becomes 0 that is undefined.
So, domain is all real numbers except -2
(b) f(x)= 2x
In this function, there is no denominator or square root or log function
so there is no restriction for x, hence domain is all real numbers
(c) f(x) = 2x-2
In this function, there is no denominator or square root or log function
so there is no restriction for x, hence domain is all real numbers
f(x) = 1/ sqrt(x+2)
if we have square root in the denominator then we set the denominator >0 and solve for x. because square root of negative values are undefined
x+2>0, x>-2
Hence domain is all real numbers that are greater than -2
Answer:
Step-by-step explanation:
If you meant " | ", as in |x|, that's "absolute value." The domain of this function is "all real numbers," and the range is "all real numbers zero or greater."
