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The curve with equation $y = f(x)$ passes through the point $(-1, 0)$ - Edexcel - A-Level Maths Pure - Question 9 - 2011 - Paper 2

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Question 9

The-curve-with-equation-$y-=-f(x)$-passes-through-the-point-$(-1,-0)$-Edexcel-A-Level Maths Pure-Question 9-2011-Paper 2.png

The curve with equation $y = f(x)$ passes through the point $(-1, 0)$. Given that $f'(x) = 12x^2 - 8x + 1$ find $f(x)$.

Worked Solution & Example Answer:The curve with equation $y = f(x)$ passes through the point $(-1, 0)$ - Edexcel - A-Level Maths Pure - Question 9 - 2011 - Paper 2

Step 1

Step 1: Integrate the derivative

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Answer

To find f(x)f(x), we start by integrating the derivative:
f(x)=(12x28x+1)dxf(x) = \int (12x^2 - 8x + 1) \, dx
This gives:
f(x)=4x34x2+x+cf(x) = 4x^3 - 4x^2 + x + c
where cc is the constant of integration.

Step 2

Step 2: Use the point $(-1, 0)$

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Answer

Next, we use the fact that the curve passes through the point (1,0)(-1, 0):
f(1)=0f(-1) = 0
Substituting x=1x = -1:
0=4(1)34(1)2+(1)+c0 = 4(-1)^3 - 4(-1)^2 + (-1) + c
This simplifies to:
0=441+c0 = -4 - 4 - 1 + c
Combining terms yields:
0=9+c0 = -9 + c
From this, we find:
c=9c = 9.

Step 3

Step 3: Write the final form of $f(x)$

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Answer

Substituting cc back into the expression for f(x)f(x), we get:
f(x)=4x34x2+x+9f(x) = 4x^3 - 4x^2 + x + 9.

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