The curve C has equation y = f(x), x ≠ 0, and the point P(2, 1) lies on C - Edexcel - A-Level Maths Pure - Question 9 - 2007 - Paper 2
Question 9
The curve C has equation y = f(x), x ≠ 0, and the point P(2, 1) lies on C. Given that
f'(x) = 3x^2 - 6 - \frac{8}{x^2},
a) find f(x).
b) Find an equation for the ... show full transcript
Worked Solution & Example Answer:The curve C has equation y = f(x), x ≠ 0, and the point P(2, 1) lies on C - Edexcel - A-Level Maths Pure - Question 9 - 2007 - Paper 2
Step 1
find f(x)
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Answer
To find f(x), we need to integrate f'(x).
Start with the given derivative:
f′(x)=3x2−6−x28
Integrate each term:
f(x)=∫(3x2)dx−∫6dx−∫x28dxf(x)=x3−6x−8⋅(−x1)+C
This simplifies to:
f(x)=x3−6x+x8+C
To find C, we use the point P(2, 1) which lies on the curve:
1=(2)3−6(2)+28+C1=8−12+4+C1=0+C
Thus, C = 1.
Therefore, the function is:
f(x)=x3−6x+x8+1
Step 2
Find an equation for the tangent to C at the point P
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Answer
To find the equation of the tangent line at point P(2, 1):
Differentiate f(x) to find f'(x):
f′(2)=3(2)2−6−(2)28=12−6−2=4
So, the gradient (m) at the point P is 4.
Use the point-slope form of the line: y−y1=m(x−x1)
Substitute P(2, 1) and m:
y−1=4(x−2)
