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A curve with equation $y = f(x)$ passes through the point (2, 10) - Edexcel - A-Level Maths Pure - Question 8 - 2012 - Paper 1

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Question 8

A-curve-with-equation-$y-=-f(x)$-passes-through-the-point-(2,-10)-Edexcel-A-Level Maths Pure-Question 8-2012-Paper 1.png

A curve with equation $y = f(x)$ passes through the point (2, 10). Given that \( f'(x) = 3x^2 - 3x + 5 \) find the value of $f(0)$.

Worked Solution & Example Answer:A curve with equation $y = f(x)$ passes through the point (2, 10) - Edexcel - A-Level Maths Pure - Question 8 - 2012 - Paper 1

Step 1

Integrate the derivative function $f'(x)$

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Answer

To find the original function f(x)f(x), we need to integrate the given derivative:
f(x)=(3x23x+5)dxf(x) = \int (3x^2 - 3x + 5) \, dx
This gives us:
f(x)=x332x2+5x+cf(x) = x^3 - \frac{3}{2}x^2 + 5x + c
where cc is the constant of integration.

Step 2

Use the point (2, 10) to find $c$

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Answer

Since the curve passes through the point (2, 10), we can substitute x=2x = 2 and f(2)=10f(2) = 10 into the equation:
10=(2)332(2)2+5(2)+c10 = (2)^3 - \frac{3}{2}(2)^2 + 5(2) + c
Calculating:
10=86+10+c10 = 8 - 6 + 10 + c
Thus,

\Rightarrow c = 10 - 12 = -2 $$. Therefore, we have $c = -2$.

Step 3

Substitute $c$ back into $f(x)$

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Answer

Now that we have cc, we substitute it back into our function:
f(x)=x332x2+5x2f(x) = x^3 - \frac{3}{2}x^2 + 5x - 2.

Step 4

Evaluate $f(0)$

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Answer

To find f(0)f(0), substitute x=0x = 0:
f(0)=(0)332(0)2+5(0)2=2f(0) = (0)^3 - \frac{3}{2}(0)^2 + 5(0) - 2 = -2.

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