3. Let f(x, y) be a function, and let (To, yo) be a nondegenerate critical point of f: it is a critical point fz(TO, yo) = fu(zo, yo) = 0, and the Hessian fax(To, yo) fwu (To, yo) - fry(zo, yo)2 is nonzero. If v = (p, q) is a nonzero vector, consider the restriction of the function f(x, y) to the line through (To, yo) with direction v: bv (t) = f(zo + tp, yo + tq). Using the chain rule, show that the derivative o'(0) vanishes, and that "(0) = fax(To, yo)p2 + 2fry(To, yo)pq + fuu(To, yo)q? = h(p, q). If fax(To, yo) fyy(To, yo) - fry(To, yo)? > 0, the derivative test for f states that the critical point (To, yo) is a local extremum. The derivative test uses the sign of frx(To, yo) to distinguish between these two cases. Show that fyy(To, yo) has the same sign as fix(To, yo) in this case, so it doesn't matter whether we use far or fyy- (In fact, the directional derivative h(p, q) is positive for all choices of p and q in this case, and not just p = 1 and q = 0, or p = 0 and q = 1.)