Question

3x^2-5x-2

Answer 1

what is the question??

Answer 2

IS THIS WHAT YOUR LOOKING FOR?preliminary: y = ax^2 + bx + c , so in your case a = 3, b = 5, c =-2
delta = b^2 - 4ac, so in your case delta = 25 - 4*3*(-2) = 25 + 24 = 49

first of all this function hasn't got an inverse because it is not bijective for the whole real set. and because of that we will have to work for two separate intervals x <or= -5/6 and x >or= -5/6 , where -5/6 is the x-coordinate of vertex of the parabola that represents the graph of the function. the vertex has the coordinates (-b/2a , -delta/4a).

i'll solve for you the inverse for the interval: x <or= -5/6
3x^2 + 5x - 2 = y
3x^2 + 5x - 2 -y = 0
3x^2 + 5x - 1*(2 +y) = 0 [solve the ecuation in x, using the formula for delta]
delta = 5^2 - 4*3*(-1)*(2+y) = 25 + 12*(2+y)
the first root is: x = [-b - sqr(delta) ] / 2a = [-5 - sqr( 25 - 12*(2+y) )] / 6
the second root is: x = [-b + sqr(delta) ] / 2a = [-5 + sqr( 25 + 12*(2+y) )] / 6
sqr means sqaure root.

because we have the condition x <or= -5/6 we'll pick the first root, to obtain smaller vales than -5/6

so we have x = [-5 - sqr( 25 - 12*(2+y) )] / 6
you can and should make the calculations: 25 -12*(2+y) = 25 - 24 - 12y = 1 - 12y
x = [-5 - sqr(1-12y)]/6
because it's just letters you can change x to y^(-1) and y to x, so the inverse is:
y^(-1) = [-5 - sqr(1-12x)]/6 for x <or= -5/6

for the other interval you take the second root, just be carefull that the function on R has no inverse, but only on the two separate intervals!

Source:

the source is thinking and knowing mathematics :)
if it helps you here is a source: http://www.purplemath.com/modules/invrsf...