Reverse what is done:
......................... y = 5x2 - 5 ........ 5 subtracted last, so add 5:
⇒ ............... y + 5 = 5x2 ............. multiplied by 5, so divide by 5:
⇒ ...... (y + 5) ÷ 5 = x2 ............... squared, so take square root:
⇒ ±√((y + 5) ÷ 5) = x ................. √n refers to the positive root of n, so need ±
Thus:
y = 5x2 - 5 ⇒ x = ±√((y + 5)/5)
fx=5x1 / 9
Yes.
Perhaps, using the Quadratic Equation...y=x2+5x-6Solve for x:x2+5x+(-6-y)=0Using the Quadratic Equation:x = .5 *(-5 (+-) (25 + (24+4y)).5) = -2.5 (+-) .5*(49+4y).5Substitute y for x, x for y:y = -2.5 (+-) .5*(49+4x).5
(x-3)(x-2)
ex and ln(x) are inverse functions. With this you can get 5x = eln(5^x) Therefore you can anti-differentiate this to get eln(5^x)/(ln(5x)) Which equals 5x/ln(5x)
No, f(x) is not the inverse of f(x).
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fx=5x1 / 9
The discriminant is 121.
Yes.
Perhaps, using the Quadratic Equation...y=x2+5x-6Solve for x:x2+5x+(-6-y)=0Using the Quadratic Equation:x = .5 *(-5 (+-) (25 + (24+4y)).5) = -2.5 (+-) .5*(49+4y).5Substitute y for x, x for y:y = -2.5 (+-) .5*(49+4x).5
x2 - 5x = 0Factor the left side:x (x - 5) = 0x = 0x = 5
(x-3)(x-2)
ex and ln(x) are inverse functions. With this you can get 5x = eln(5^x) Therefore you can anti-differentiate this to get eln(5^x)/(ln(5x)) Which equals 5x/ln(5x)
If f(x)=(5x-1)/9 Then the inverse is x=(5y-1)/9 9x=5y-1 9x+1=5y (9x+1)/5=y The inverse is f(x)=(9x+1)/5
That factors to (x - 2)(x + 7)