x < -1 = }2x2 + 1, -1 1 2х — 3, 17. If f(x) 6, f(-2)

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Q: det fi s→ IR be afanction and 8= [-202] 5-+ $$ f(x) = -25x2-1 -15x20 3 05341 2 1<x<2 ร

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Q: 1 If f(x, у, 2)- ôf then х— у—3z ду а. 1 (х- у-3-)2 b. -1 (х-у-3-)* С. 1 (х-у-3-)° d. 2 (х - у-3-)?

A: In Partial differentiation w.r.t y we treat x,z as constants.

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