What is the inverse and domain of f(x)=x^2+4

2 answers:

7 0

The domain of the inverse of a function is the range of the original function

so find the range of f(x)
this is lowest to highest
we got the vertex is at (0,4)
and it goes to infinity
so range is from 4 to infinity

inverse
replace f(x) with y
switch x and y
solve for y
replace y with finverse

f(x)=x^2+4
y=x^2+4
x=y^2+4
x-4=y^2
y=√(x-4)
f⁻¹(x)=√(x-4)

inverse is f⁻¹(x)=√(x-4)
the domain is x≥4

Dominik [7]3 years ago

4 0

To find the inverse let
y = x^2 + 4
x^2 = y - 4
x = sqrt(y - 4)

inverse f-1(x) = sqrt( x - 4)

the domain of x^2 + 4 is all real x

the domain of f-1(x) is x>= 4

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Two brothers want to share a large field equally.

Elan Coil [88]

Yes, the partition gives the two brothers equal shares.

Step-by-step explanation:

Step 1:

Assume the entire field has an area of B. So one brother takes $\frac{1}{12} B$ , $\frac{1}{6}B$ , and $\frac{1}{4} B$

So we need to calculate how much this brother takes in terms of B.

To do this we calculate how much $\frac{1}{12} B + \frac{1}{6} B + \frac{1}{4} B$ is.

Step 2:

To add $\frac{1}{12} B + \frac{1}{6} B + \frac{1}{4} B$ ,

First take the LCM of the denominators 12, 6, and 4

The LCM is 12, we multiply the denominator to get the LCM value, this same value is multiplied to the numerator too.

$\frac{1}{12} B + \frac{1}{6} B + \frac{1}{4} B = \frac{1(1)}{12(1)} B + \frac{1(2)}{6(2)} B + \frac{1(3)}{4(3)} B \\\\\frac{1}{12} B + \frac{1}{6} B + \frac{1}{4} B= \frac{1B +2B+3B}{12} = \frac{B}{2}$