If f(x)= -4x^2+7 and x≤0, what is the equation of the inverse function?
The answer is supposed to be f^-1(x)= -[√(7-x)]/2, but this is what I did:
x=-4y^2+7
-4y^2=x-7
y^2=(x-7)/-4
y= √[(x-7)/-4]
Did I do something wrong?
By the way, for the original function, the domain would be x≤0, but would the range be (-∞,7]?
Thanks!
To find the equation of the inverse function, you correctly started by substituting x with y and solving for y. However, a small mistake occurred when you simplified the equation.
x = -4y^2 + 7
-4y^2 = x - 7
y^2 = (x - 7) / -4
y = ±√((x - 7) / -4)
At this point, you correctly took the square root of both sides, but you missed the negative sign that comes from the square root of a negative number:
y = ±√(-(x - 7) / 4)
To eliminate the ± symbol and keep the inverse function as a single-valued function, we can drop the positive solution and only consider the negative solution:
y = -√(-(x - 7) / 4)
So, the correct equation for the inverse function is:
f^(-1)(x) = -√((7 - x) / 4)
Regarding the domain and range of the original function, you are correct that the domain would be x ≤ 0. As for the range, we can find it by analyzing the graph of the function or observing the leading coefficient (-4) of the quadratic term. Since the coefficient is negative, the graph opens downward, which means the maximum value occurs at the vertex. In this case, the vertex is (0, 7) since the x-coordinate is 0. So, the range of the function would be (-∞, 7].