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

x = - 2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x = 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = - 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $x = - 2$
The minimum value of the function exists at x=c on the condition $f^{^{'}} (c) = 0$
Now, $f = x^{3} + 2 x^{2} - 4 x + 6$
$f^{'} (x) = 3 x^{2} + 4 x - 4$
So, $f^{'} (c) = 3 c^{2} + 4 c - 4 = 0$ here c is the point where f is minimum.
$\Rightarrow 3 c^{2} + 4 c - 4 = 0$
$\Rightarrow 3 c^{2} + 6 c - 2 c - 4 = 0$
$\Rightarrow (3 c - 2) (c + 2) = 0$
$\Rightarrow c = \frac{2}{3}, - 2$
Thus, the value of c is $- 2$ .

Suggest Corrections

Similar questions

Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

(i) (ii)

(iii)

(iv)

Find the absolue maximum value and he absolute minimum value of the following functions in the given intervals:

$f (x) = x^{3}, x ϵ [- 2, 2]$

$f (x) = s i n x + c o s x, x ϵ [0, π]$

$f (x) = 4 x - \frac{1}{2} x^{2}, x ϵ [- 2, \frac{9}{2}]$

$f (x) = (x - 1)^{2} + 3, x ϵ [- 3, 1]$

If $f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} x^{3}, & x < 0 3 x - 2, & 0 \leq x \leq 2 x^{2} + 1, & x > 2 \end{matrix}$
Then find the value(s) of x for which f(x)=2.

Q. Consider the function for

x = [- 2, 3]

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{x^{3} - 2 x^{2} - 5 x + 6}{x - 1}, & i f x \neq 1 - 6 & i f x = 1 \end{matrix}

then

Examinie the continuity of the function $f (x) = x^{3} + 2 x^{2} - 1$ at $x = 1$

Join BYJU'S Learning Program

The maximum value of the function f(x)=x3+2x2−4x+6 exists at

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