1) Find the value of the lim(x->0) ((sin5x/sin2x)-(sin3x-4x)). I have no idea where to start so if someone could start it off for me it would really help.
as x-->0, sin x --> x
(5x/2x - 3x + 4x)
5/2
1) Find the value of the lim(x->0) ((sin5x/sin2x)-(sin3x-4x)). I have no idea where to start so if someone could start it off for me it would really help.
2) If f(x) = (2x-3)/(5x+4) then what's the inverse of f(x)
So I replaced x with y
x = (2y-3)/(5y-3)
Now am I supposed to isolate y? Can someone help me out with that.
3) If f(x)=arcsin(x/3), then what's f'(sq5).
I set f'(x) = arcsinx/arcsin3..am I on the right path?
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4) Let f(x) = ax^2+bx, where a and b are constants. If the tangent line to the curve y=f(x) at the point (1,1) has equation y = 3x-2, then what's f(3)?
What I did was first find f(1)
and I got a+b
Then I took the derivative of f'(x) and got 2a+b..so I then took the derivative of 2(1)+(1) and got 3 which is the slope.
Therefore 2a+b = 3. How do I find a and b?
6 answers
2) If f(x) = (2x-3)/(5x+4) then what's the inverse of f(x)
So I replaced x with y
x = (2y-3)/(5y-3)
Now am I supposed to isolate y? Can someone help me out with that.
5 y x - 3 x = 2 y - 3
5 y x - 2 y = 3 x - 3
y (5 x - 2) = 3 (x-1)
y = 3(x-1)/(5x-2)
So I replaced x with y
x = (2y-3)/(5y-3)
Now am I supposed to isolate y? Can someone help me out with that.
5 y x - 3 x = 2 y - 3
5 y x - 2 y = 3 x - 3
y (5 x - 2) = 3 (x-1)
y = 3(x-1)/(5x-2)
3) If f(x)=arcsin(x/3), then what's f'(sq5).
I set f'(x) = arcsinx/arcsin3..am I on the right path?
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d/dx (sin^-1 u ) = du/dx /sqrt(1-u^2)
here u = x/3
so
f'(x) =(1/3) / sqrt (1 - x^2/9)
now f'(sqrt 5)
f'(sqrt 5) = (1/3) / sqrt (1 - 5/9)
= (1/3) / sqrt (4/9)
= 1/2
I set f'(x) = arcsinx/arcsin3..am I on the right path?
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d/dx (sin^-1 u ) = du/dx /sqrt(1-u^2)
here u = x/3
so
f'(x) =(1/3) / sqrt (1 - x^2/9)
now f'(sqrt 5)
f'(sqrt 5) = (1/3) / sqrt (1 - 5/9)
= (1/3) / sqrt (4/9)
= 1/2
the answer to this question is OMG THAT ISHARD JESE KID WAT GARDE R U IN
woops i mean this........ the answer to this question is..... OMG KID THAT IS HARD, WHAT GRADE R U IN
4) Let f(x) = ax^2+bx, where a and b are constants. If the tangent line to the curve y=f(x) at the point (1,1) has equation y = 3x-2, then what's f(3)?
What I did was first find f(1)
and I got a+b
Then I took the derivative of f'(x) and got 2a+b..so I then took the derivative of 2(1)+(1) and got 3 which is the slope.
Therefore 2a+b = 3. How do I find a and b?
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Then I took the derivative of f'(x) and got 2a+b..
I GET 2 a x + b but that does not matter because where we are x = 1
so 2 a + b = 3 so b = 3 - 2 a
so
f(x) = a x^2 + (3-2a)x
but it goes through (1,1)
1 = a + 3 - 2 a
a = 2
I think you can take it from there
What I did was first find f(1)
and I got a+b
Then I took the derivative of f'(x) and got 2a+b..so I then took the derivative of 2(1)+(1) and got 3 which is the slope.
Therefore 2a+b = 3. How do I find a and b?
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Then I took the derivative of f'(x) and got 2a+b..
I GET 2 a x + b but that does not matter because where we are x = 1
so 2 a + b = 3 so b = 3 - 2 a
so
f(x) = a x^2 + (3-2a)x
but it goes through (1,1)
1 = a + 3 - 2 a
a = 2
I think you can take it from there
2 answers