Find a function f(x) such that the point (-1,1) is on the graph of y=f(x), the slope of the tangent line at (-1,1) is 2 and f” (x) = 6x + 4?
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To find a function ( f(x) ) that satisfies the given conditions, we can follow these steps:
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Since the point ((-1, 1)) is on the graph of ( y = f(x) ), we know that ( f(-1) = 1 ).
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The slope of the tangent line at ((-1, 1)) is given as 2. This means that the derivative of ( f(x) ) at ( x = -1 ) is ( f'(x) = 2 ).
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We are given the second derivative of ( f(x) ), ( f''(x) = 6x + 4 ).
To find ( f'(x) ), we integrate ( f''(x) ):
[ f'(x) = \int (6x + 4) , dx ]
[ f'(x) = 3x^2 + 4x + C_1 ]
Since ( f'(x) = 2 ) when ( x = -1 ), we can solve for the constant ( C_1 ):
[ 2 = 3(-1)^2 + 4(-1) + C_1 ]
[ 2 = 3 - 4 + C_1 ]
[ C_1 = 3 ]
So, ( f'(x) = 3x^2 + 4x + 3 ).
Now, to find ( f(x) ), we integrate ( f'(x) ):
[ f(x) = \int (3x^2 + 4x + 3) , dx ]
[ f(x) = x^3 + 2x^2 + 3x + C_2 ]
Since ( f(-1) = 1 ), we can solve for the constant ( C_2 ):
[ 1 = (-1)^3 + 2(-1)^2 + 3(-1) + C_2 ]
[ 1 = -1 + 2 - 3 + C_2 ]
[ C_2 = 3 ]
So, ( f(x) = x^3 + 2x^2 + 3x + 3 ).
Therefore, the function ( f(x) ) that satisfies the given conditions is ( f(x) = x^3 + 2x^2 + 3x + 3 ).
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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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