question concerns the graph of the function f(x) = 3(x + 1)^2 -12.
(i) What is the image set of the function f? Give the answer in
interval notation.
(ii)
Explain why the function f as defined above does not have an
inverse.
(i) What is the image set of the function f? Give the answer in
interval notation.
(ii)
Explain why the function f as defined above does not have an
inverse.
-
i) quick answer: just by looking at the function: Its shape is k(x - h)² + v, where (h,v) is the vertex and the sign of k is the direction the branches point.
So the vertex is located at (-1,-12). This with the fact its branches point infinitely upwards means that the vertex is located at the overall lowest point of the function.
Therefore, the image set is [-12, +infinity[
i) full-blown answer: If we differentiate the function, we can find the extremes:
f'(x) = 6(x + 1) = 6x + 6
This is a positive linear function (6 > 0), so it's positive to the right of its root and negative to the left. This means that f(x) decreases until (f')'s root and increases afterwards, and therefore that f(x) has an overall minimum at that root.
Let's find (f')'s root:
f'(x) = 6x + 6 = 0
6x = -6
x = -1
Therefore, f(x) reaches its all-time low at x = -1. f(-1) = -12, so the lower boundary of the image set is -12.
Next, we know that parabolae are defined over all |R, and reach infinity both right and left of the vertex. This means the upper boundary of the image set is +infinity.
Hence our image set of [-12,+infinity[
ii) By definition, a function that has an inverse has to be bijective. This means that for every y value in the image set, there is one AND ONLY ONE x value that gives that particular value. This is not the case here, as for example, for y = 0, you have:
3(x+1)² - 12 = 0
3(x+1)² = 12
(x+1)² = 4
x+1 = 2 or -2
x = 1 or -3, which is two values.
So the vertex is located at (-1,-12). This with the fact its branches point infinitely upwards means that the vertex is located at the overall lowest point of the function.
Therefore, the image set is [-12, +infinity[
i) full-blown answer: If we differentiate the function, we can find the extremes:
f'(x) = 6(x + 1) = 6x + 6
This is a positive linear function (6 > 0), so it's positive to the right of its root and negative to the left. This means that f(x) decreases until (f')'s root and increases afterwards, and therefore that f(x) has an overall minimum at that root.
Let's find (f')'s root:
f'(x) = 6x + 6 = 0
6x = -6
x = -1
Therefore, f(x) reaches its all-time low at x = -1. f(-1) = -12, so the lower boundary of the image set is -12.
Next, we know that parabolae are defined over all |R, and reach infinity both right and left of the vertex. This means the upper boundary of the image set is +infinity.
Hence our image set of [-12,+infinity[
ii) By definition, a function that has an inverse has to be bijective. This means that for every y value in the image set, there is one AND ONLY ONE x value that gives that particular value. This is not the case here, as for example, for y = 0, you have:
3(x+1)² - 12 = 0
3(x+1)² = 12
(x+1)² = 4
x+1 = 2 or -2
x = 1 or -3, which is two values.
-
(i) (x+1)^2 is never negative and thus can only be greater than or equal to zero...
(x+1)^2 >= 0
3(x+1)^2 >= 0
3(x+1)^2 - 12 >= -12
f(x) >= -12
Thus, the image set is [12, INFINITY[
(ii) f(x) does not have an inverse since there could be 2 values for x that map to the same image f(x)
i.e., f(x) is not one-to-one
f(0) = f(-2) = -9
f(1) = f(-3) = 0
(x+1)^2 >= 0
3(x+1)^2 >= 0
3(x+1)^2 - 12 >= -12
f(x) >= -12
Thus, the image set is [12, INFINITY[
(ii) f(x) does not have an inverse since there could be 2 values for x that map to the same image f(x)
i.e., f(x) is not one-to-one
f(0) = f(-2) = -9
f(1) = f(-3) = 0