Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 4 PROBLEMS PLUS ¤ 229
To put this in terms of m and b, wesolvethesystemy = x 2 1 and y = mx 1 + b, givingusx 2 1 − mx 1 − b =0
⇒
√
x 1 = 1 2 m − m2 +4b √
. Similarly, x 2 = 1 2 m + m2 +4b . The area is then 1 (x √
8 2 − x 1 ) 3 = 1 m2 +4b 3
8 ,
and is attained at the point P
x P ,x 2 P = P 1 m, 1 m2 .
2 4
Note: Another way to get an expression for f(x) is to use the formula for an area of a triangle in terms of the coordinates of
the vertices: f(x) = 1 2 x2x 2 1 − x 1x2 2 + x1x 2 − xx1 2 + xx
2
2 − x 2x 2 .
15. Suppose that the curve y = a x intersects the line y = x. Thena x 0
= x 0 for some x 0 > 0, and hence a = x 1/x 0
0 .Wefind the
maximum value of g(x) =x 1/x , > 0, because if a is larger than the maximum value of this function, then the curve y = a x
does not intersect the line y = x. g 0 (x) =e (1/x)lnx − 1 x ln x + 1 2 x · 1 1
= x 1/x (1 − ln x). Thisis0 only where
x
x 2
x = e,andfor0 e, f 0 (x) < 0,sog has an absolute maximum of g(e) =e 1/e .Soif
y = a x intersects y = x,wemusthave0