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A curve has equation $y = f(x)$ and passes through the point $(4, 22)$ - Edexcel - A-Level Maths Pure - Question 6 - 2009 - Paper 1

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A curve has equation $y = f(x)$ and passes through the point $(4, 22)$. Given that $f'(x) = 3x^2 - 3x^2 - 7$, use integration to find $f(x)$, giving each term ... show full transcript

Worked Solution & Example Answer:A curve has equation $y = f(x)$ and passes through the point $(4, 22)$ - Edexcel - A-Level Maths Pure - Question 6 - 2009 - Paper 1

Step 1

Integration of $f'(x)$

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Answer

To find f(x)f(x), we need to integrate f′(x)f'(x):

f(x)=∫(3x2−3x−7) dxf(x) = \int (3x^2 - 3x - 7) \, dx

This can be solved term by term:

f(x)=3x33−3x22−7x+c=x3−32x2−7x+cf(x) = \frac{3x^3}{3} - \frac{3x^2}{2} - 7x + c = x^3 - \frac{3}{2}x^2 - 7x + c

Step 2

Finding the constant $c$ using point (4, 22)

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Answer

We use the given point (4,22)(4, 22) to find the constant cc:

Substituting x=4x = 4 into f(x)f(x):

f(4)=43−32(42)−7(4)+c=22f(4) = 4^3 - \frac{3}{2}(4^2) - 7(4) + c = 22

Calculating:

64−32(16)−28+c=2264 - \frac{3}{2}(16) - 28 + c = 22

This simplifies to:

64−24−28+c=2264 - 24 - 28 + c = 22

Thus:

12+c=2212 + c = 22

Therefore:

c=10c = 10

Step 3

Final form of $f(x)$

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Answer

Substituting the value of cc back into the equation:

f(x)=x3−32x2−7x+10f(x) = x^3 - \frac{3}{2}x^2 - 7x + 10

This is the required function in its simplest form.

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