help please dont know this!?

Question

Let 

Give the largest domain on which  is one-to-one and non-decreasing.    

Give the range of  .    

Find the inverse of  restricted to the domain above.

   

Give the domain of  .   

Give the range of  .  

Answer 1

Nico

If you plot this on a graphing calculator or Desmos you will see it is always 1-1 (for a given x, one value of y)

The function is zero when x = 7, so touches the x axis at (7,0). To the left of (7,0) funcon is decreasing (as x increases, y decreases), to the right of (7,0) the function is increasing.

so the domain (x values) where f is increasing is x >7 or (7, +∞)

Range of f (possible y values) is [0,+∞)

for the inverse

f (x) = (x-7)²

lets put f(x) = y

y = (x-7) ²

to find the inverse function

1. get x in terms of y
2. switch the x and y

y = (x-7)²

√y = x - 7

7 + √y = x

switch x, y

7 + √x = y

y = f^-1 (x)

f^-1 (x) = 7 + √x

Domain of the inverse : f^-1 (x) will exist as long as x >= 0, (so the square root exists) so the domain should be [0, + ∞). However the question states the inverse is restricted to the domain above, so domain is x > 7 or (7, +∞).

Range of the inverse. Obviously is is strictly increasing (as x increases , y increases) so the minimum values must be 7 + √7, and maximum + ∞. So range is ( 7 + √7, + ∞)

MIKE

Answer 2

I wish we could insert graphs or anything more than plain text, but I will try to help.

(a) Find the root(s) of the function, the zeros. This one is simple (x-7)^2 = 0 only when x=7. That means the curve, we know it's a parabola because of the squared, touches the x-axis once at x=7. Since the x in our f(x) is positive, the parabola curves upwards. Hence, all the points from 7 to + infinity are non-decreasing.

Alternatively, you could take the derivative of (x-7)^2 to get 2*(x-7) which is positive or non-zero when x >=7. The on-to-one part just means that every x produces a unique value of y. Notice that both x = 11 and x=3 give the same value y = 16?

(b) The range is the possible outputs of the function. Because of the squared and there's no vertical shift, f(x) can never be negative. Thus, the range is [0, infinity).

(c) To get the inverse, write y=(x-7)^2 then switch y and x to get x=(y-7)^2 and solve for y again.

x = (y-7)^2 →√x = y-7 → y = √x + 7. This is fine on the domain [7, infinity).

(d) The domain is the set of x-values that produce real values of y, I'm assuming we are concerned only with Real numbers. The only restriction then on the inverse is that x has to be greater than or equal to zero, it can't be negative. So, the domain is [0,infinity).

(e) If you plot the inverse, it's easy to see that the range is [7, infinity). However, just put in the lower value for the domain into the equation for x and you'll get y = -7 which is the minimum value of y.