1.
a)
f(x)=ln(−2x2−x−6)+x2−1
−2x2−x−6>0x2−1≥0
D=(−1)2−4(−2)(−6)=−47<0 Then −2x2−x−6<0,x∈R
Domain:{}
b)
f(x)=ln(−x2−x−62)+x2−1
x2−x−6<0x2−1≥0
(x+2)(x−3)<0
x≤−1 or x≥1 Domain:(−2,−1]∪[1,3)
2.
f(x)=x2−4x+5,x∈[3,4]
xv=−2(1)−4=2 The function f increases on (3,4)
f(3)=(3)2−4(3)+5=2
f(4)=(4)2−4(4)+5=5Domain: [3,4]
Range: [2,5]
y=x2−4x+5,3≤x≤4 Change x and y
x=y2−4y+5,3≤y≤4Solve for y
y2−4y+4=x−1(y−2)2=x−1 Since 3≤y≤4
y−2=x−1
y=2+x−1
Then
f−1(x)=2+x−1
Domain: [2,5]
Range: [3,4]
3.
f(x)=4xx−2x Domain: [0,∞)
f′(x)=4(23)x−2x2=x6x−1
slope=f′(x)=x6x−1=1
6x−x−1=0
(3x+1)(2x−1)=0 Since x≥0, we take 2x−1=0
x=21
x=41
f(41)=4(41)41−241=−21The tangent line to the graph is
y+21=x−41
y=x−43
4.
f(x)=ex+3x2+2x+6 Domain: (−∞,∞)
f′(x)=ex+6x+2
f′(x)=0=>ex+6x+2=0
(ex+6x+2)′=ex+6>0,x∈RThe function g(x)=ex+6x+2 is strictly increasing on (−∞,∞).
The equation ex+6x+2=0 has the only solution x≈−0.4406.
The function f(x) is monotonic decreasing on (−∞,−0.4406).
The function f(x) is monotonic increasing on (−0.4406,∞).
5.
∫6x3ex2+2dx t=x2,dt=2xdx
∫6x3ex2+2dx=3e2∫tetdt
∫tetdt
u=t,du=dt
dv=etdt,v=et
∫tetdt=tet−∫etdt=tet−et+C1
∫6x3ex2+2dx=3ex2+2(x2−1)+C
6.
x2+1+y′cos(y)=0
cos(y)dy=−(x2+1)dx Integrate
∫cos(y)dy=−∫(x2+1)dx
sin(y)=−3x3−x+C
sin(y)+3x3+x=C
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