(a)
f(x)=(−2x+1)2x2
−2x+1=0=>x=1/2D(f):(−∞,1/2)∪(1/2,∞)
x→(1/2)−lim(−2x+1)2x2=∞
x→(1/2)+lim(−2x+1)2x2=∞ Vertical asymptote: x=1/2
x→−∞lim(−2x+1)2x2=x→−∞lim(−2x/x+1/x)2x2/x2
=x→−∞lim(−2+1/x)21=1/4
x→∞lim(−2x+1)2x2=x→∞lim(−2x/x+1/x)2x2/x2
=x→∞lim(−2+1/x)21=1/4 Horizontal asymptote: y=1/4
There is no slant (oblique) asymptote.
(b)
f′(x)=((−2x+1)2x2)′
=(−2x+1)42x(−2x+1)2−x2(2)(−2)(−2x+1)
=(−2x+1)3−4x2+2x+4x2=(−2x+1)32x
Find critical number(s)
f′(x)=0=>(−2x+1)32x=0
x=0 (i)
If x<0,f′(x)<0,f(x) decreases.
If 0<x<1/2,f′(x)>0,f(x) increases.
If x>1/2,f′(x)<0,f(x) decreases.
The function f increases on (0,1/2).
The function f decreases on (−∞,0)∪(1/2,∞).
(ii)
The function f has a local minimum at x=0.
The function f has no a local maximum.
(c)
f′′(x)=((−2x+1)32x)′
=2⋅(−2x+1)6(−2x+1)3−x(3)(−2)(−2x+1)2
=2⋅(−2x+1)4−2x+1+6x=(−2x+1)48x+2 Find the inflection point(s)
f′′(x)=0=>(−2x+1)48x+2=0
x=−1/4 (i)
If x<−1/4,f′′(x)<0,f(x) is concave down.
If −1/4<x<1/2,f′′(x)>0,f(x) is concave up.
If x>1/2,f′′(x)>0,f(x) is concave up.
The function f is concave up on (−1/4,1/2)∪(1/2,∞).
The function f is concave down on (−∞,−1/4).
(ii)
The function f has inflection point at x=−1/4.
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