Question

The function \(f(x)=4x^3-18x^2+27x-7\) has:-

• A. Maxima at \(x=\frac{3}{2}\)
• B. Minima at \(x=\frac{3}{2}\)
• C. Point of inflexion at \(x=\frac{3}{2}\)
• D. Maximum value of function is 156

Hint:

\(f'(x)=3(2x-3)^2\) never changes sign, and \(f''(3/2)=0\) with \(f'''\ne0\).

Answer 1

The Correct Option is C

Step 1: Find the critical points.
Differentiate and set the derivative to zero.
\[ f'(x)=12x^2-36x+27=3(4x^2-12x+9)=3(2x-3)^2 \] So \(f'(x)=0\) only at \(x=\frac32\).

Step 2: Look at the sign of f'.
Since \(3(2x-3)^2\ge 0\) for every \(x\), the derivative does not change sign at \(x=\frac32\). It is positive on both sides. A maximum or minimum needs a sign change, so there is no extremum there.

Step 3: Use the second derivative.
\[ f''(x)=24x-36 \Rightarrow f''\left(\tfrac32\right)=36-36=0 \] The test fails, so check the third derivative: \(f'''(x)=24\ne 0\). A non-zero third derivative where \(f''=0\) means a point of inflexion.

Step 4: Check the options.
Option 1 (maxima) and option 2 (minima) are wrong because \(f'\) does not change sign. Option 4 is wrong because a cubic with positive leading coefficient goes to \(+\infty\), so it has no maximum value. Option 3 is correct.

Final Answer:
The function has a point of inflexion at \(x=\frac32\), option 3. \[ \boxed{\text{Point of inflexion at } x=\tfrac32} \]