What is the equation of the line tangent to #f(x)=2x-x^2 # at #x=0#?
y = 2x
To find the equation of the tangent : y - b = m(x-a)
Require to know m , the gradient and (a , b ) a point on the line.
Now, m = f'(x) and a is given in the question (x = 0 ). To find b ,
substitute x = 0 into f(x)
f'(x) = 2 - 2x → f'(0) = 2 = m.
and f(0) = 0 → (a , b ) = (0 , 0 )
This is a line passing thro' the origin with m =2 and hence
equation is y = 2x
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The equation of the line tangent to f(x)=2x-x^2 at x=0 is y = 2x.
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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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