NCERT Class 12 Maths · Chapter 6 Applications Of Derivatives · Miscellaneous Exercise Question 10

Question

For f(x)=(x−2)⁴(x+1)³ find (i) local maxima, (ii) local minima, (iii) points of inflexion.

Accepted Answer

Final Answer: Local maximum at x=2/7; local minimum at x=2. Inflections at x=−1 and (2±3√2)/7; points are (x,f(x)). Step 1: f′=(x−2)³(x+1)²(7x−2). Its sign is positive up to 2/7, negative between 2/7 and 2, and positive after 2; the isolated zero at −1 does not change this sign. Step 2: Thus x=2/7 is a local maximum, x=2 a local minimum, and x=−1 is neither. Step 3: f″=6(x−2)²(x+1)(7x²−4x−2). Its zeros are −1, 2, (2−3√2)/7, (2+3√2)/7. Step 4: The three simple roots −1 and (2±3√2)/7 each change the sign of f″, hence are inflections. The double root 2 does not. Step 5: The corresponding inflection coordinates are ($-1$, $0$); ($\frac{2}{7} - \frac{3 \sqrt{2}}{7}$, $- \frac{2187 \left(-3 + \sqrt{2}\right)^{3} \left(\sqrt{2} + 4\right)^{4}}{823543}$); ($\frac{2}{7} + \frac{3 \sqrt{2}}{7}$, $\frac{2187 \left(-4 + \sqrt{2}\right)^{4} \left(\sqrt{2} + 3\right)^{3}}{823543}$).

NCERT Class 12 Maths · Chapter 6 Applications Of Derivatives · Miscellaneous Exercise Question 10

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