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?
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
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)
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
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
1) To find the value of the limit as x approaches 0, let's break it down step by step:
Start with the expression: lim(x->0) ((sin5x/sin2x)-(sin3x-4x))
Step 1: Simplify each term separately:
For the first term, sin(5x)/sin(2x), you can use the limit property that states lim(x->0) (sin(x)/x) = 1. Applying this property, you have:
lim(x->0) (sin(5x)/(5x)) / (sin(2x)/(2x)) = 5/2.
For the second term, sin(3x) - 4x, you can directly evaluate it at x=0, which gives you:
lim(x->0) (sin(3x) - 4x) = sin(0) - 4(0) = 0.
Step 2: Combine the simplified terms:
lim(x->0) ((sin5x/sin2x)-(sin3x-4x)) = (5/2) - 0 = 5/2.
Therefore, the value of the limit is 5/2.
2) To find the inverse of the function f(x) = (2x-3)/(5x+4), you can follow these steps:
Step 1: Replace f(x) with y:
y = (2x-3)/(5x+4).
Step 2: Swap x and y:
x = (2y-3)/(5y+4).
Step 3: Solve for y:
Multiply both sides by (5y+4) to eliminate the denominator:
x(5y+4) = 2y-3.
Expand and rearrange:
5xy + 4x = 2y - 3.
Move all y-terms to one side and all x-terms to the other side:
5xy - 2y = -4x - 3.
Factor out y:
y(5x - 2) = -4x - 3.
Divide both sides by (5x - 2):
y = (-4x - 3)/(5x - 2).
So, the inverse of f(x) is f^(-1)(x) = (-4x - 3)/(5x - 2).
3) To find f'(sqrt(5)) given f(x) = arcsin(x/3), you need to use the chain rule. Here are the steps:
Step 1: Differentiate the function f(x) = arcsin(x/3) with respect to x:
f'(x) = 1/sqrt(1 - (x/3)^2).
Step 2: Evaluate the derivative at x = sqrt(5):
f'(sqrt(5)) = 1/sqrt(1 - (sqrt(5)/3)^2).
Simplify further:
f'(sqrt(5)) = 1/sqrt(1 - (5/9)) = 1/sqrt(4/9) = 1/(2/3) = 3/2.
Therefore, f'(sqrt(5)) = 3/2.
4) To find the values of a and b in the function f(x) = ax^2 + bx, given that the tangent line at (1,1) has the equation y = 3x - 2, follow these steps:
Step 1: Find f(1):
Substitute x = 1 into the function f(x) = ax^2 + bx:
f(1) = a(1)^2 + b(1) = a + b.
Step 2: Find f'(x):
Take the derivative of f(x) = ax^2 + bx with respect to x:
f'(x) = 2ax + b.
Step 3: Evaluate f'(1):
Substitute x = 1 into f'(x):
f'(1) = 2a(1) + b = 2a + b.
Step 4: Set up the equation for the tangent line:
Given that the tangent line at (1,1) has the equation y = 3x - 2, we have the following conditions:
f(1) = 1,
f'(1) = 3.
Step 5: Solve the system of equations:
From Step 1: a + b = 1,
From Step 3: 2a + b = 3.
Solving this system of equations, you can use various methods, such as substitution or elimination, to find the values of a and b.
Once you find the values of a and b, you can substitute them back into the original function f(x) = ax^2 + bx and evaluate f(3) by substituting x = 3.