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: Rewrite f with a shifted variable.
Let \(t=x-\frac32\). Then \(2x-3=2t\) and \(f'(x)=3(2t)^2=12t^2\). Integrating gives \(f=4t^3+C\), so the graph looks like \(t^3\) shifted up or down.

Step 2: Recall the shape of t cubed.
The graph of \(t^3\) is always rising and flattens at \(t=0\), where it crosses from curving down to curving up. That point is a point of inflexion.

Step 3: Confirm with f'' changing sign.
\(f''(x)=24x-36=24t\). This is negative for \(t<0\) and positive for \(t>0\). So the curve changes concavity at \(x=\frac32\).

Step 4: Value at that point.
\(f\left(\frac32\right)=4\cdot\frac{27}{8}-18\cdot\frac94+27\cdot\frac32-7=13.5-40.5+40.5-7=6.5\). It is not 156, and being a rising cubic it has no largest value either.

Final Answer:
So \(x=\frac32\) is a point of inflexion. \[ \boxed{\text{Point of inflexion at } x=\tfrac32} \]