for -2 < x < +3 and f(x) = 3x^2 - 6x
a)determine the domain and range of f(x)
b) the eqn of f inverse if f(x) is further restricted to 1 < x < 3.
i don't know how to these problems i tried everything please help me
Sure, I can help you with these problems! Let's tackle them one by one.
a) To determine the domain and range of the function f(x) = 3x^2 - 6x, we first need to identify the possible values of x within the given range of -2 < x < 3.
The domain refers to the set of possible input values for the function. In this case, since there are no restrictions on the values of x, the domain is the entire real number line (-∞, +∞).
Now let's find the range, which refers to the set of possible output values. To determine the range, we should analyze the behavior of the function. In this case, f(x) is a quadratic function where the leading coefficient is positive (3), so the parabola opens upward.
To find the vertex (the maximum or minimum point) of the parabola, we can use the formula x = -b/2a, where a is the coefficient of x^2 (-6 in this case) and b is the coefficient of x (-6 in this case). Plugging these values into the formula, we get x = -(-6) / (2 * 3) = 1.
Now, substitute x = 1 back into the function f(x) to get f(1) = 3(1)^2 - 6(1) = 3 - 6 = -3. So the vertex of the parabola is at (1, -3).
Since the parabola opens upward and the vertex is the minimum point, we know that the range of the function is from the y-value at the vertex (-3) to positive infinity. Therefore, the range of f(x) is (-∞, -3].
b) Now, let's find the equation of the inverse function of f(x) when the domain is further restricted to 1 < x < 3.
To find the inverse of f(x), we need to interchange the roles of x and y. So let's start by replacing f(x) with y in the original equation:
y = 3x^2 - 6x
Next, we'll swap x and y:
x = 3y^2 - 6y
Now, let's solve this equation for y to find the inverse:
x = 3y^2 - 6y
Rearrange the terms:
3y^2 - 6y - x = 0
Now, solve this quadratic equation for y. We can use the quadratic formula: y = (-b ± √(b^2 - 4ac)) / (2a).
In this case, a = 3, b = -6, and c = -x. Plugging these values into the quadratic formula, we get:
y = (-(-6) ± √((-6)^2 - 4 * 3 * (-x))) / (2 * 3)
y = (6 ± √(36 + 12x)) / 6
y = (1 ± √(9 + 3x)) / 3
This equation represents the inverse of f(x) when the domain is further restricted to 1 < x < 3.