find the inverse of the function 3x^2+x+10
asked by guest on Nov 13, 2024 at 11:44 pm
You asked:
MathBot Answer:
For −∞<x≤−16-\infty \lt x \le - \frac{1}{6}−∞<x≤−61 the inverse is −12x−1196−16- \frac{\sqrt{12 x - 119}}{6} - \frac{1}{6}−612x−119−61For −16≤x<∞- \frac{1}{6} \le x \lt \infty−61≤x<∞ the inverse is 12x−1196−16\frac{\sqrt{12 x - 119}}{6} - \frac{1}{6}612x−119−61
