Class 12 Applications Of Derivatives · Long Answer3 Markslocal_extrema

Class 12 Maths · Applications Of Derivatives · Long Answer #99

Find all local maxima and minima of f(x)=3x⁴/4−8x³−45x²/2+105.

✅ Step-by-Step Solution

Final Answer: Local maximum (0,105); local minima at x=4±√31 with f(x)=−5183/4∓248√31.

Differentiate: f′=3x³−24x²−45x=3x(x²−8x−15).
Its zeros are α=4−√31<0, 0 and β=4+√31>0. The derivative signs are −,+,−,+ across these ordered roots.
Thus α and β are local minima and 0 is a local maximum. Substituting x=4±√31 (using x²=8x+15) gives f(4±√31)=−5183/4∓248√31.
The local maximum is f(0)=105; the two minima occur at 4−√31 and 4+√31 with the stated values.

[0.5 marks] Differentiate: f′=3x³−24x²−45x=3x(x²−8x−15).
[1 mark] Its zeros are α=4−√31<0, 0 and β=4+√31>0. The derivative signs are −,+,−,+ across these ordered roots.
[1 mark] Thus α and β are local minima and 0 is a local maximum. Substituting x=4±√31 (using x²=8x+15) gives f(4±√31)=−5183/4∓248√31.
[0.5 marks] The local maximum is f(0)=105; the two minima occur at 4−√31 and 4+√31 with the stated values.

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