Q214 / 270 · NDA 2013 (Phase 2) · Mathematics

The maximum value of the function $f(x)=x^3+2 x^2-$ $4 x+6$ exists at

  • Correct answer
    A
    x = –2
  • B
    x = 1
  • C
    x = 2
  • D
    x = –1

Solution

$\begin{aligned} & f(x)=x^3+2 x^2-4 x+6 \\ & f^{\prime}(x)=3 x^2+4 x-4 \end{aligned}$ For extremum values of $f(x), f^{\prime}(x)=0$ $3 x^2+4 x-4 =0$ $\begin{aligned}(3 x-2)(x+2) & =0 \Rightarrow x=-2, \frac{2}{3} \\ f^{\prime \prime}(x) & =6 x+4\end{aligned}$ $\begin{aligned} {for \quad}x=-2, \quad f^{\prime \prime}(-2) & =6(-2)+4 \\ & =-8 \lt 0\end{aligned}$ Hence, $f(x)$ is maximum at $x=-2$.

Hence, the correct answer is option: 'A'.

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