How do you find the points that the line tangent to #f(x)=x/(x+2)# has a slope 1/2?
The two points are (0,0) and (-4,2).
Known fact- f'(x)/gradient = 1/2.
Thought process- what are some possible values of (x,y) when f'(x) = 1/2?
f(x) = 1 - 2/(x+2) (By long division)
When x = 0 f(0) = 0
when x = -4 f(-4) = 2
The two points are (0,0) and (-4,2).
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To find the points where the line tangent to the function f(x) = x/(x+2) has a slope of 1/2, we need to find the derivative of the function and set it equal to 1/2.
First, we find the derivative of f(x) using the quotient rule:
f'(x) = [(x+2)(1) - x(1)] / (x+2)^2
Next, we set the derivative equal to 1/2 and solve for x:
[(x+2)(1) - x(1)] / (x+2)^2 = 1/2
Simplifying the equation, we get:
2(x+2) - 2x = (x+2)^2
Expanding and simplifying further:
2x + 4 - 2x = x^2 + 4x + 4
Rearranging the terms:
x^2 + 4x + 4 - 2x - 2x - 4 = 0
Simplifying:
x^2 - 2x = 0
Factoring out x:
x(x - 2) = 0
Setting each factor equal to zero:
x = 0 or x - 2 = 0
Solving for x, we get:
x = 0 or x = 2
Therefore, the points where the line tangent to f(x) = x/(x+2) has a slope of 1/2 are (0, 0) and (2, 2/4).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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