The maximum value of $x^{3} - 3 x$ in the interval [0, 2] is

-2

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

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

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

Open in App

Solution

The correct option is B 2
$f (x) = x^{3} - 3 x$
$f^{'} (x) = 3 (x^{2} - 1)$
For maxima or minima,
$f^{'} (x) = 0$
$\Rightarrow x = - 1, 1$
$\Rightarrow x = 1$ (since -1 does not belongs in the given interval)
$f " (x) = 6 x$
$f^{''} (1) = 6 > 0$ , so f(x) has a minimum at $x = 1$ .
Now we will find the value of function at end points
$f (0) = 0$
$f (2) = 8 - 6 = 2$
So, maximum value is 2.

Suggest Corrections

Similar questions

x

x^{2} - 3 x + 2 > 0

x^{2} - 3 x - 4 \leq 0

f (x) = x^{3} - 6 x^{2} + 9 x + 1

x = 0 to x = 5

f (x) = x^{3} - 6 x^{2} + 9 x + 1

x = 0 to x = 5

\sqrt{3}

f (x) = x^{3} - 9 x^{2} + 24 x + 5

[1, 6]

Join BYJU'S Learning Program