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f(x) = \frac{-3}{x + 2} + \frac{3}{(x + 2)^2},\ x \neq -2 - Edexcel - A-Level Maths Pure - Question 4 - 2007 - Paper 6
Question 4
f(x) = \frac{-3}{x + 2} + \frac{3}{(x + 2)^2},\ x \neq -2. (a) Show that \( f(x) = \frac{x^2 + x + 1}{(x + 2)^2} \),\ x \neq -2. (b) Show that \( x^2 + x + 1 > 0 \... show full transcript
Worked Solution & Example Answer:f(x) = \frac{-3}{x + 2} + \frac{3}{(x + 2)^2},\ x \neq -2 - Edexcel - A-Level Maths Pure - Question 4 - 2007 - Paper 6
Step 1
Show that \( f(x) = \frac{x^2 + x + 1}{(x + 2)^2} \),\ x \neq -2.
Answer
We start with the function:
To combine these fractions, we find a common denominator, which is ( (x + 2)^2 ). We rewrite the first term as:
Thus, we have:
Expanding the numerator, we get:
So,
From the simplification, we see that:
Step 2
Show that \( x^2 + x + 1 > 0 \) for all values of \( x \).
Answer
To demonstrate that ( x^2 + x + 1 > 0 ), we can analyze the discriminant of the quadratic equation.
For the expression ( x^2 + x + 1 ), we identify the coefficients: ( a = 1 ), ( b = 1 ), and ( c = 1 ).
The discriminant ( D ) is given by:
Since the discriminant is negative, the quadratic does not intersect the x-axis and is always positive. Therefore,
Step 3
Show that \( f(x) > 0 \) for all values of \( x,\ x \neq -2. \)
Answer
From part (b), we established that ( x^2 + x + 1 > 0 ). Now, since the denominator ( (x + 2)^2 ) is also positive for all ( x \neq -2 ) (as it is a square and doesn't equal zero), we can conclude:
Thus, we show that ( f(x) > 0 ) holds true under the given conditions.