f(x) = x(4 - x) The range is f(x)<= 4 How is this worked out?

Question

f(x) = x(4 - x) The range is f(x)<= 4 How is this worked out?

Answer 1

This is a parabola that opens downward. Find the vertex. The y-value of the vertex will be the largest value that f(x) can take.
f(x) = x(4-x)
Distribute the x.
f(x) = 4x - x^2
f(x) = -x^2 + 4x (You can tell the parabola opens downward because the number in front of x^2 is negative).
The vertex occurs at x = -b/(2a), where a is the coefficient of x^2 and b is the coefficient of x.
a = -1 and b = 4
x = -4/(2*-1)
x = -4/-2
x = 2
To find y, replace x with 2
y = f(2) = -(2)^2 + 4(2)
y = -4 + 8
y = 4
Thus, the vertex is at (2,4). Since it opens downward, the range is all values less than or equal to 4.

Answer 2

Are you asking how you find the range of the function?

It's what the possible f(x) values are for the function. You can plot it to find the range.

Or you could take the derivative of the function and find the extremum and then put it back in to the function.

f'(x) = 4 -2x

The extemum is at x= 2, and f'(x) is decreasing so x=2 is a maximum.

So, f(2) = 4. f(x) <=4.

Answer 3

f(x) = x(4-x) cannot have a value > 4.
Tou can find this out by trial-and error, or by finding the maximum of the function:
f(x) = 4x-x^2
df/dx = 4 - 2x = 0 for max or min
2x = 4
x = 2
so, at x = 2, you have f(x) = 2(4-2) = 4
We know that this is a maximum, because f(x) is a parabola which opens "downward".
Therefore the range of f(x) <= 4

Answer 4

f(x) = x(4 - x)
=4x-x^2,
which is a parabola that opens down.
so its maximum is the highest y can get:
f'(x) = 4 -2x =, when x= 2, it is the only critical point and corresponds to the maximum:
f(2) = 4 and f(x) <=4 for all other x's

Answer 5

f(x)=x(4-x)
f(x)=4x-x^2

Vertex is (2,4)...y=a(x-p)^2+q ---> y=-(x-2)^2+4

Parabela opens down (a value is negative) so, y values less than vertex...
f(x)<=4

Hope that solved your question