IB Math Analysis & Approaches Question bank-Topic SL 5.1 Introduction to the concept of a limit SL Paper 1

Detailed solution

(a) Find the Derivatives

(i) \(\frac{dy}{dx}\)

To find the first derivative, we differentiate \( y = x^3 – x^2 – x + 1 \) term by term:
– Derivative of \( x^3 \) is \( 3x^2 \) (power rule: \( \frac{d}{dx}[x^n] = n x^{n-1} \)),
– Derivative of \( -x^2 \) is \( -2x \),
– Derivative of \( -x \) is \( -1 \),
– Derivative of \( 1 \) is \( 0 \).

So:
\[
\frac{dy}{dx} = 3x^2 – 2x – 1
\]
This gives us the slope of the tangent at any point

(ii) \(\frac{d^2 y}{dx^2}\)

Now, the second derivative comes from differentiating \(\frac{dy}{dx} = 3x^2 – 2x – 1\):
– Derivative of \( 3x^2 \) is \( 6x \),
– Derivative of \( -2x \) is \( -2 \),
– Derivative of \( -1 \) is \( 0 \).

Thus:
\[
\frac{d^2 y}{dx^2} = 6x – 2
\]

(b) Find the Coordinates of the Local Maximum A

A local maximum occurs where the slope is zero (first derivative equals 0) and the curve is concave down (second derivative is negative). Start with \(\frac{dy}{dx} = 0\):
\[
3x^2 – 2x – 1 = 0
\]
Solve this quadratic using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \), where \( a = 3 \), \( b = -2 \), \( c = -1 \):
– Discriminant: \( b^2 – 4ac = (-2)^2 – 4 \cdot 3 \cdot (-1) = 4 + 12 = 16 \),
 \( x = \frac{-(-2) \pm \sqrt{16}}{2 \cdot 3} = \frac{2 \pm 4}{6} \).

Two solutions:
 \( x = \frac{2 + 4}{6} = \frac{6}{6} = 1 \),
 \( x = \frac{2 – 4}{6} = \frac{-2}{6} = -\frac{1}{3} \).

We have two critical points. To determine which is the local maximum, use the second derivative test with \(\frac{d^2 y}{dx^2} = 6x – 2\):
– At \( x = 1 \): \( 6 \cdot 1 – 2 = 6 – 2 = 4 > 0 \) (concave up, so a local minimum),
– At \( x = -\frac{1}{3} \): \( 6 \left(-\frac{1}{3}\right) – 2 = -2 – 2 = -4 < 0 \) (concave down, so a local maximum).

The local maximum is at \( x = -\frac{1}{3} \). Now compute \( y \):
\[
y = \left(-\frac{1}{3}\right)^3 – \left(-\frac{1}{3}\right)^2 – \left(-\frac{1}{3}\right) + 1
\]
 \( \left(-\frac{1}{3}\right)^3 = -\frac{1}{27} \),
 \( \left(-\frac{1}{3}\right)^2 = \frac{1}{9} \),
 \( -\left(-\frac{1}{3}\right) = \frac{1}{3} \),
 Combine: \( y = -\frac{1}{27} – \frac{1}{9} + \frac{1}{3} + 1 \).

Use a common denominator (27):
 \( -\frac{1}{27} = -\frac{1}{27} \),
 \( -\frac{1}{9} = -\frac{3}{27} \),
 \( \frac{1}{3} = \frac{9}{27} \),
 \( 1 = \frac{27}{27} \),
 Total: \( -\frac{1}{27} – \frac{3}{27} + \frac{9}{27} + \frac{27}{27} = \frac{-1 – 3 + 9 + 27}{27} = \frac{32}{27} \).

Coordinates of A are \(\left(-\frac{1}{3}, \frac{32}{27}\right)\). The negative second derivative confirms the maximum

(c) Find the x-Coordinate of the Point of Inflection B

A point of inflection occurs where the concavity changes, so \(\frac{d^2 y}{dx^2} = 0\):
\[
6x – 2 = 0
\]
\[
6x = 2 \quad \Rightarrow \quad x = \frac{2}{6} = \frac{1}{3}
\]
Check concavity change:
– For \( x < \frac{1}{3} \), e.g., \( x = 0 \): \( 6 \cdot 0 – 2 = -2 < 0 \) (concave down),
– For \( x > \frac{1}{3} \), e.g., \( x = 1 \): \( 6 \cdot 1 – 2 = 4 > 0 \) (concave up).

The second derivative changes sign at \( x = \frac{1}{3} \), confirming an inflection point. We only need the x-coordinate, so \( x = \frac{1}{3} \).

(d) Find the Equation of the Normal Line at (0, 1)

The normal line is perpendicular to the tangent at \( (0, 1) \). First, the slope of the tangent:
\[
\frac{dy}{dx} = 3x^2 – 2x – 1
\]
At \( x = 0 \): \( 3 \cdot 0^2 – 2 \cdot 0 – 1 = -1 \).

Tangent slope is \(-1\), so the normal slope is the negative reciprocal: \( -\frac{1}{-1} = 1 \). Using point-slope form at \( (0, 1) \):
\[
y – 1 = 1 (x – 0)
\]
\[
y – 1 = x \quad \Rightarrow \quad y = x + 1
\]

………………………Markscheme…………………………

Ans:

(a)     (i) \(\frac{dy}{dx}=3x^{2}-2x-1\)

(ii) \(\frac{d^{2}y}{dx^{2}}=6x-2\)

(b) setting their (quadratic) \(\frac{dy}{dx}\) to 0 and solve using valid method

substituting one of their x values into \(\frac{d^{2}y}{dx^{2}}\)

EITHER

at \(x=-\frac{1}{3}\),  \(\frac{d^{2}y}{dx^{2}}=6\left ( -\frac{1}{3} \right )-2(=-4)\) < 0 so local max

OR

at x = 1, \(\frac{d^{2}y}{dx^{2}}=6(1)-2(=4)\) > 0 so local min hence local max at \(x=-\frac{1}{3}\)

THEN

substituting their x-coordinate of A into y

\(y=\left ( -\frac{1}{3} \right )^{3}-\left ( -\frac{1}{3} \right )^{2}-\left ( -\frac{1}{3} \right )+1\)

\(y=\frac{32}{27}\)

so coordinates of A are \(\left ( -\frac{1}{3}, \frac{32}{27} \right )\)

(c) setting their \(\frac{d^{2}y}{dx^{2}}\) to 0

\(6x-2=0\)

\(x=\frac{1}{3}\)

(d)  gradient of tangent = \(-1\)

        negative reciprocal of their gradient

        gradient of normal = \(1\)

       equation is \(y=x+1\)  (accept point/slope form \(y-1=(x-0)\))