One to one means only one x value gives one y value
This only happens when x is greater than or equal to -1 [-1, infinity) or when x is less than or equal to -1 (-infinity, -1].
Non decreasing means the slope is zero or positive, so it has to be the first option for the domain [-1, infinity]
The inverse of f(x) = (x + 1)^2
y = (x + 1)^2
x = (y + 1)^2
Sq rt x = y + 1
(sq rt x) - 1 = y
f^-1(x) = (sq rt x) - 1
We cannot take the square root of -1, so the domain of the inverse of f is [0, infinity]
Answer:
5x5 is 25 and 5+5+5+5+5 is also 25 please give brainliest
Step-by-step explanation:
Answer: idk explanation: I’m dumb
<span>1.) d - 1 > -3 and d + 2 < 5
d-1>-3→d-1+1>-3+1→d>-2
</span>d+2<5→d+2-2<5-2→d<3
<span>Part 1: The solution is d>-2 and d<3.
Part 2: There would be an open circle on -2 and an open circle on 3.
Part 3: The shading is between the two numbers.
</span><span>
2.) -3k + 7 > -5 and -7k + 5 < -2
-3k+7>-5→-3k+7-7>-5-7→-3k>-12→(-1/3)(-3k>-12)→k<4
-7k+5<-2→-7k+5-5<-2-5→-7k<-7→(-1/7)(-7k<-7)→k>1
</span>Part 1: The solution is k>1 and k<4.
Part 2: There would be an open circle on 1 and an open circle on 4.
Part 3: The shading is between the two numbers.<span>
3.)-5 < 2y - 3 < 23 (Hint: refer to example 2 in the lesson)
-5<2y-3<23→-5+3<2y-3+3<23+3→-2<2y<26→-2/2<2y/2<26/2→-1<y<13
</span>Part 1: The solution is y>-1 and y<13.
Part 2: There would be an open circle on -1 and an open circle on 13.
Part 3: The shading is between the two numbers.<span>
4.)x - 4 < 5 or x + 2 > 15
x-4<5→x-4+4<5+4→x<9
x+2>15→x+2-2>15-2→x>13
</span>Part 1: The solution is x>9 or x>13.
Part 2: There would be an open circle on 9 and an open circle on 13.
Part 3: The shading is from 9 to the left and from 13 to the right.<span>
5.) 5x - 4 > 6 or -4x + 5 > 1
5x-4>6→5x-4+4>6+4→5x>10→5x/5>10/5→x>2
-4x+5>1→-4x+5-5>1-5→-4x>-4→(-1/4)(-4x>-4)→x<1
</span>Part 1: The solution is x<1 or x>2.
Part 2: There would be an open circle on 1 and an open circle on 2.
Part 3: The shading is from 1 to the left and from 2 to the right.