Find the range and domain of followingfunctions
f(x) = whole root(2 + x - x2)
Answers
Answered by
0
f(x)=√2+x-x²
let f(X) = y
y=√2+x-x²
squaring both sides
y²=2+x-x²
y² is always a positive number so it means that f(X)≥0
2+x-x²≥0
multiply the whole equation by (-)1
x²-x-2≤0 because inquality sign changes
factorise it
x²-2x+x-2≤0
x(x-2)+1(x-2)≤0
(x+1)(x-2)≤0
x=-1and +2
so there are three cases
- infinity + -1 - 2 + +infinity
↑-----------------↑----------------↑-----------------↑
mean domain of this function is [-1,2]
--------------------------------------------------
for range of the function
x²-x-2≤0
by completing square method
(x²-x/2 - 1/4 -2 ≤ 0
(x-1/2)² + -1/4 -2 ≤0
(x-1/2)²≤9/4
x-1/2≤+-3/2
x≤+3/2+1/2=2
and x≤-3/2+1/2=-1
so minimum value of this function is -1 and maximum value of this function is 2 so range of the fuction ∈[-1 ,2]
let f(X) = y
y=√2+x-x²
squaring both sides
y²=2+x-x²
y² is always a positive number so it means that f(X)≥0
2+x-x²≥0
multiply the whole equation by (-)1
x²-x-2≤0 because inquality sign changes
factorise it
x²-2x+x-2≤0
x(x-2)+1(x-2)≤0
(x+1)(x-2)≤0
x=-1and +2
so there are three cases
- infinity + -1 - 2 + +infinity
↑-----------------↑----------------↑-----------------↑
mean domain of this function is [-1,2]
--------------------------------------------------
for range of the function
x²-x-2≤0
by completing square method
(x²-x/2 - 1/4 -2 ≤ 0
(x-1/2)² + -1/4 -2 ≤0
(x-1/2)²≤9/4
x-1/2≤+-3/2
x≤+3/2+1/2=2
and x≤-3/2+1/2=-1
so minimum value of this function is -1 and maximum value of this function is 2 so range of the fuction ∈[-1 ,2]
Similar questions