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quadratic functions help please (1 Viewer)
- Thread starter chazza
- Start date
1. This is not a quadratic functions question - it is a calculus / curve sketching problem.chazza said:
2. The approach needed is:
If k = xe<sup>x/2</sup> has two different roots, then the curves y = k and y = xe<sup>x/2</sup> must meet exactly twice. y = k is a horizontal line, so you need to sketch y = xe<sup>x/2</sup> to solve this problem.
3. y = xe<sup>x/2</sup> has:
Domain of all real x
Only intercept at (0, 0)
It is neither odd nor even.
As x ---> +∞, y ---> +∞
As x ---> -∞, y ---> 0<sup>-</sup>
dy/dx = (x + 2)e<sup>x/2</sup> / 2, from which it follows that it has a minimum turning point at (-2, -2 / e)
d<sup>2</sup>y/dx<sup>2</sup> = (x + 4)e<sup>x/2</sup> / 4, from which it follows that it has an inflexion at (-4. -4 / e<sup>2</sup>)
From this, it is not difficult to sketch y = xe<sup>x/2</sup>
4. Looking at your sketch, it should now be obvious that a horizontal line y = k will meet the curve
once if k > 0
once if k = 0
twice if -2 / e < k < 0
once if k = -2 / e
never if k < -2 /e
Thus, k = xe<sup>x/2</sup> has two distinct roots only if -2 / e < k < 0