1.
a)
f(x)=ln(โ2x2โxโ6)+x2โ1โ
โ2x2โxโ6>0x2โ1โฅ0
D=(โ1)2โ4(โ2)(โ6)=โ47<0 Then โ2x2โxโ6<0,xโR
Domain:{}
b)
f(x)=ln(โx2โxโ62โ)+x2โ1โ
x2โxโ6<0x2โ1โฅ0
(x+2)(xโ3)<0
xโคโ1 or xโฅ1 Domain:(โ2,โ1]โช[1,3)
2.
f(x)=x2โ4x+5,xโ[3,4]
xvโ=โ2(1)โ4โ=2 The function f increases on (3,4)
f(3)=(3)2โ4(3)+5=2
f(4)=(4)2โ4(4)+5=5Domain: [3,4]
Range: [2,5]
y=x2โ4x+5,3โคxโค4 Change x and y
x=y2โ4y+5,3โคyโค4Solve for y
y2โ4y+4=xโ1(yโ2)2=xโ1 Since 3โคyโค4
yโ2=xโ1โ
y=2+xโ1โ
Then
fโ1(x)=2+xโ1โ
Domain: [2,5]
Range: [3,4]
3.
f(x)=4xxโโ2xโ Domain: [0,โ)
fโฒ(x)=4(23โ)xโโ2xโ2โ=xโ6xโ1โ
slope=fโฒ(x)=xโ6xโ1โ=1
6xโxโโ1=0
(3xโ+1)(2xโโ1)=0 Since xโโฅ0, we take 2xโโ1=0
xโ=21โ
x=41โ
f(41โ)=4(41โ)41โโโ241โโ=โ21โThe tangent line to the graph is
y+21โ=xโ41โ
y=xโ43โ
4.
f(x)=ex+3x2+2x+6 Domain: (โโ,โ)
fโฒ(x)=ex+6x+2
fโฒ(x)=0=>ex+6x+2=0
(ex+6x+2)โฒ=ex+6>0,xโRThe function g(x)=ex+6x+2 is strictly increasing on (โโ,โ).
The equation ex+6x+2=0 has the only solution xโโ0.4406.
The function f(x) is monotonic decreasing on (โโ,โ0.4406).
The function f(x) is monotonic increasing on (โ0.4406,โ).
5.
โซ6x3ex2+2dx t=x2,dt=2xdx
โซ6x3ex2+2dx=3e2โซtetdt
โซtetdt
u=t,du=dt
dv=etdt,v=et
โซtetdt=tetโโซetdt=tetโet+C1โ
โซ6x3ex2+2dx=3ex2+2(x2โ1)+C
6.
x2+1+yโฒcos(y)=0
cos(y)dy=โ(x2+1)dx Integrate
โซcos(y)dy=โโซ(x2+1)dx
sin(y)=โ3x3โโx+C
sin(y)+3x3โ+x=C
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