Question
The derivative of a function $f$ is given by $f'(x) = \frac{2x+2}{x^2+2x+2}$, for $x \in \mathbb{R}$.
(a) (i) Show that $x^2+2x+2 > 0$ for all values of $x$.
(ii) Hence, find the values of $x$ for which $f$ is increasing.
(b) (i) Write down the value of $x$ for which $f'(x) = 0$.
(ii) Show that $f”(x) = \frac{-2x^2-4x}{(x^2+2x+2)^2}$.
(iii) Hence, justify that the value of $x$ found in part (b)(i) corresponds to a local minimum point on the graph of $f$.
It is given that $f(2) = 3 + \ln 10$.
(c) Find an expression for $f(x)$.
(d) Find the equation of the normal to the graph of $f$ at $(2, 3 + \ln 10)$.
▶️Answer/Explanation
Detailed solution
(a)(i) To show \( x^2 + 2x + 2 > 0 \) for all values of \( x \)
The expression \( x^2 + 2x + 2 \) appears in the denominator of the derivative \( f'(x) = \frac{2x + 2}{x^2 + 2x + 2} \), so we need to determine its sign.
consider it as a quadratic \( x^2 + 2x + 2 \):
Discriminant: \( \Delta = b^2 – 4ac = 2^2 – 4 \cdot 1 \cdot 2 = 4 – 8 = -4 < 0 \).
Since \( a = 1 > 0 \) and the discriminant is negative, the quadratic has no real roots and opens upward, so it’s always above the x-axis.
Thus, \( x^2 + 2x + 2 > 0 \) for all \( x \in \mathbb{R} \).
(a)(ii) Hence, find the values of \( x \) for which \( f \) is increasing
A function \( f \) is increasing where its derivative \( f'(x) > 0 \). Given \( f'(x) = \frac{2x + 2}{x^2 + 2x + 2} \):
From (a)(i), the denominator \( x^2 + 2x + 2 > 0 \) for all \( x \).
The sign of \( f'(x) \) depends on the numerator \( 2x + 2 \).
Solve \( 2x + 2 > 0 \):
\( 2x + 2 > 0 \)
\( 2x > -2 \)
\( x > -1 \).
Since the denominator is always positive, \( f'(x) > 0 \) when \( x > -1 \). Let’s check:
If \( x = 0 > -1 \): \( f'(0) = \frac{2 \cdot 0 + 2}{0 + 0 + 2} = \frac{2}{2} = 1 > 0 \).
If \( x = -2 < -1 \): \( f'(-2) = \frac{2(-2) + 2}{4 – 4 + 2} = \frac{-4 + 2}{2} = \frac{-2}{2} = -1 < 0 \).
So, \( f \) is increasing for \( x > -1 \).
(b)(i) Write down the value of \( x \) for which \( f'(x) = 0 \)
Set the derivative equal to zero:
\( f'(x) = \frac{2x + 2}{x^2 + 2x + 2} = 0 \).
Since the denominator is never zero (from (a)(i)), set the numerator to zero: \( 2x + 2 = 0 \).
\( 2x = -2 \)
\( x = -1 \).
Thus, \( f'(x) = 0 \) at \( x = -1 \).
(b)(ii) Show that \( f”(x) = \frac{-2x^2 – 4x}{(x^2 + 2x + 2)^2} \)
Differentiate \( f'(x) = \frac{2x + 2}{x^2 + 2x + 2} \) using the quotient rule. Let:
\( u = 2x + 2 \), so \( u’ = 2 \),
\( v = x^2 + 2x + 2 \), so \( v’ = 2x + 2 \),
Quotient rule: \( f”(x) = \frac{u’ v – u v’}{v^2} \).
Compute:
Numerator: \( u’ v – u v’ = 2 (x^2 + 2x + 2) – (2x + 2) (2x + 2) \).
Expand:
\( 2 (x^2 + 2x + 2) = 2x^2 + 4x + 4 \),
\( (2x + 2)(2x + 2) = (2x + 2)^2 = 4x^2 + 8x + 4 \),
\( 2x^2 + 4x + 4 – (4x^2 + 8x + 4) = 2x^2 + 4x + 4 – 4x^2 – 8x – 4 = -2x^2 – 4x \).
Denominator: \( v^2 = (x^2 + 2x + 2)^2 \).
So, \( f”(x) = \frac{-2x^2 – 4x}{(x^2 + 2x + 2)^2} \), as required.
(b)(iii) Justify that \( x = -1 \) corresponds to a local minimum
At \( x = -1 \) (from (b)(i)):
\( f'(-1) = 0 \) (critical point).
Evaluate \( f”(-1) \):
Numerator: \( -2(-1)^2 – 4(-1) = -2(1) + 4 = -2 + 4 = 2 \),
Denominator: \( ((-1)^2 + 2(-1) + 2)^2 = (1 – 2 + 2)^2 = 1^2 = 1 \),
\( f”(-1) = \frac{2}{1} = 2 > 0 \).
Since \( f”(-1) > 0 \), the second derivative test indicates a local minimum at \( x = -1 \).
(c) Find an expression for \( f(x) \)
Integrate \( f'(x) = \frac{2x + 2}{x^2 + 2x + 2} \):
Notice \( x^2 + 2x + 2 = (x + 1)^2 + 1 \) (from (a)(i)).
Rewrite the numerator: \( 2x + 2 = 2(x + 1) \).
So, \( f'(x) = \frac{2(x + 1)}{(x + 1)^2 + 1} \).
This suggests a substitution: let \( u = x + 1 \), \( du = dx \),
Then \( f'(x) = \frac{2u}{u^2 + 1} \), and \( \int \frac{2u}{u^2 + 1} du = \ln(u^2 + 1) + C \) (since the derivative of \( \ln(u^2 + 1) \) is \( \frac{2u}{u^2 + 1} \)).
Back-substitute: \( f(x) = \ln((x + 1)^2 + 1) + C = \ln(x^2 + 2x + 2) + C \).
Use \( f(2) = 3 + \ln 10 \):
\( f(2) = \ln(2^2 + 2 \cdot 2 + 2) + C = \ln(4 + 4 + 2) + C = \ln 10 + C \),
Set equal: \( \ln 10 + C = 3 + \ln 10 \),
\( C = 3 \).
Thus, \( f(x) = 3 + \ln(x^2 + 2x + 2) \).
(d) Find the equation of the normal at \( (2, 3 + \ln 10) \)
Slope of tangent: \( f'(2) = \frac{2 \cdot 2 + 2}{2^2 + 2 \cdot 2 + 2} = \frac{6}{10} = \frac{3}{5} \).
Normal slope = negative reciprocal = \( -\frac{5}{3} \).
Point-slope form: \( y – (3 + \ln 10) = -\frac{5}{3} (x – 2) \).
Simplify:
\( y – 3 – \ln 10 = -\frac{5}{3} x + \frac{10}{3} \),
\( y = -\frac{5}{3} x + \frac{10}{3} + 3 + \ln 10 = -\frac{5}{3} x + \frac{19}{3} + \ln 10 \).
Final equation: \( y = -\frac{5}{3} x + \frac{19}{3} + \ln 10 \).
………………..Markscheme…………………
Solution:-
(a) (i) METHOD 1
$x^2+2x+2 = (x+1)^2+1$
$(x+1)^2 \geq 0$
$(x+1)^2+1 > 0$
METHOD 2
discriminant $\Delta = 4-8 = (-4)$
$\Delta < 0$ and concave up OR $\Delta < 0$ and coefficient of $x^2>0$ (may be seen in diagram)
hence $x^2+2x+2>0$
(ii) $x>-1$ (accept $x\geq-1$)
(b) (i) x = -1
(ii) attempt to use the quotient rule
$f'(x) = \frac{2(x^2+2x+2) – (2x+2)(2x+2)}{(x^2+2x+2)^2}$
$f'(x) = \frac{2x^2+4x+4-4x^2-8x-4}{(x^2+2x+2)^2}$ OR $f'(x) = \frac{2x^2+4x+4-(4x^2+8x+4)}{(x^2+2x+2)^2}$
$f'(x) = \frac{-2x^2-4x}{(x^2+2x+2)^2}$
(iii) substituting x = -1 into f'(x), clearly leading to positive numerator
$f'(-1) = \frac{-2+4}{1} = (2)$
$f'(-1) > 0$
therefore this is a local minimum point