IB Mathematics AHL 5.10 The second derivative-AI HL Paper 1- Exam Style Questions- New Syllabus

The function \(f\) is defined by

\(f(x)=\dfrac{6}{2x^2+5x+4}\).

(a) Find \(f'(x)\). [2]

(b) Hence or otherwise, find the \(x\)-coordinates of the points of inflexion of \(f\). [3]

Most-appropriate topic codes (IB DP Mathematics: Applications and Interpretation HL):

• TOPIC AHL 5.9 The chain rule, product rule and quotient rule for differentiation. (Part a)

• TOPIC AHL 5.10 The second derivative, concavity and points of inflexion. (Part b)

▶️ Answer/Explanation

(a)

Write the function using a negative power:

\(f(x)=6(2x^2+5x+4)^{-1}\).

Using the chain rule:

\(f'(x)=6(-1)(2x^2+5x+4)^{-2}(4x+5)\).

Therefore,

\(f'(x)=-6(4x+5)(2x^2+5x+4)^{-2}\).

Equivalently,

✅ \(f'(x)=\dfrac{-6(4x+5)}{(2x^2+5x+4)^2}\)

(b)

A point of inflexion occurs where the concavity changes. We therefore find the second derivative and solve \(f”(x)=0\).

Differentiating \(f'(x)\) gives:

\(f”(x)=\dfrac{12(12x^2+30x+17)}{(2x^2+5x+4)^3}\).

Since

\(2x^2+5x+4>0\)

for all real \(x\), the second derivative is zero when its numerator is zero:

\(12x^2+30x+17=0\).

Using the quadratic formula:

\(x=\dfrac{-30\pm\sqrt{30^2-4(12)(17)}}{2(12)}\).

\(x=\dfrac{-30\pm\sqrt{84}}{24}\).

\(x=-\dfrac{5}{4}\pm\dfrac{\sqrt{21}}{12}\).

Therefore,

\(x=-1.631881\ldots\)

\(x=-0.868118\ldots\).

Both roots are distinct, and \(f”(x)\) changes sign at each value, so both correspond to points of inflexion.

✅ Answer: \(x\approx-1.63\) and \(x\approx-0.868\)

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