f(x) = square root of [(8x^2 +1)/ (9-5x^2)]
I am so confused and I have a test on monday please help..
2007-10-27 10:32:55 · 2 answers · asked by Slava T 1 in Science & Mathematics ➔ Mathematics
Range is [1/3, .6sqrt(5))
y is imaginary if 9-5x^2)<0 which is when x > sqrt(9/5)
y^2 = (8x^2+1)/(9-5x^2)
9y^2 -5x^2y^2 = 8x^2 +1
9y^2 - 1 = 8x^2 +5 x^2y^2 = x^2(8+5y^2)
x^2 = (9y^2-1)/(8+y^2)
x = sqrt[(9y^2-1)/(8+y^2)], or interchanging x and y,
y = sqrt[(9x^2-1)/(8+x^2)] <-- inverse
2007-10-27 10:51:37 · answer #1 · answered by ironduke8159 7 · 0⤊ 0⤋
ironduke got most of it right but he dropped the 5 in his work
the correct inverse f'(x) is equal to sq.rt of (9x^2-1)/(5x^2+8)
the domain of f is -.6sqrt(5) less than x less than .6sqrt(5)
the range of f is the same as the domain of f'
the domain of f' is x less than -1/3 or x greater than 1/3
the range of f therefore is y must be less than -1/3 or greater than 1/3
2007-10-27 11:27:03 · answer #2 · answered by Anonymous · 0⤊ 0⤋