∫(x3+1)2(3x2)dx
1/2x6+x3+C
y=lnx dx
1/x
∫1/x(dx)
lnx+C
For a function to have an inverse, it must be?
1:1 function
Solve, 50e-x=30
x≈0.511
∫√(2x+1)dx
1/3(2x+1)2/3+C
y=ln(4x)dx
1/x
∫1/(4x)dx
(1/4)lnx + C
A graphical test that can be used to determine if a function has an inverse is the what?
The horizontal line test
Find the derivative, y=e5x
5e5x
∫e5xdx
1/5e5x+C
y=ln√(3x+1)
3/(6x+2)
1/3ln (3x+1) + C
Find the inverse, f(x)=x2
f-1(x)=√x
Find the derivative, y=(x+1)2ex
ex(x+1)(x+3)
∫√(2x+1)dx from 0 to 4
26/3
y=ln(x−2)−ln(x+2)
ln((x-2)/(x+2))
∫(x2-2)/(x+1)dx, [0,2]
-ln3
Find f-1(7), f(x)=5-2x3
-1/6
Evaluate, ∫e5x-3dx
(1/5)e5x-3+C
∫dx/(x√lnx) from e to e4
2
x2−3lny+y2 =10
(2xy)/(3-2y2)
Avg. value of f(x)=8/x2, [2,4]
1
True or False? If f is an even function, then f-1 exists
False
∫e-2xdx, [0,1]
(e2-1)/2e2