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?

Answer 1

#f(x)=x^3+2x^2+3x+4#

We have #f''(x)=6x+4#
#f'(x)=int(6x+4)dx=3x^2+4x+c_1#
As slope of tangent at #(-1,1)# is #2#,
#f'(-1)=3*(-1)^2+4*(-1)+c_1=2# or #3-4+c_1=2# i.e. #c_1=3#
Hence #f'(x)=3x^2+4x+3#
and #f(x)=int(3x^2+4x+3)dx=x^3+2x^2+3x+c_2#
Now point #(-1,1)# lies on this curve, hence
#f(-1)=(-1)^3+2*(-1)^2+4*(-1)+c_2=1#
or #-1+2-4+c_1=1# i.e. #c_1=1+1-2+4=4#
Hence #f(x)=x^3+2x^2+3x+4#
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Answer 2

To find a function ( f(x) ) that satisfies the given conditions, we can follow these steps:

  1. Since the point ((-1, 1)) is on the graph of ( y = f(x) ), we know that ( f(-1) = 1 ).

  2. 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 ).

  3. 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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Answer from HIX Tutor

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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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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