The minimum values of $f (x) = 2^{(x^{2} - 3)^{3} + 27}$

2^{27}

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

1

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

4

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

16

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

Open in App

Solution

The correct option is D $1$
$f (x) = 2^{(x^{2} - 3)^{3} + 27}$
$\Rightarrow f (x) = 2^{(x^{6} - 9 x^{4} + 27 x^{2})}$
$f^{'} (x) = (6 x^{5} - 36 x^{3} + 54 x) 2^{(x^{6} - 9 x^{4} + 27 x^{2})} l o g_{e} 2$
For maxima or minima,
$f^{'} (x) = 0$
$6 x^{5} - 36 x^{3} + 54 x = 0$
$x = 0$ , $\sqrt{3}$ ,- $\sqrt{3}$
$f^{''} (x) > 0$ at $x = 0$
Hence, f(x) has a minimum value at $x = 0$
$f (0) = 1$

Suggest Corrections

Similar questions

2^{(x^{2} - 3)^{3} + 27}

x + \frac{1}{x} (x > 0)

4 \times 2^{(x^{2} - 3)^{3} + 27}

5 {sin}^{2} x + 3 {cos}^{2} x

The minimum value of f(x) = -2 $x^{2}$ + 5x + 4 ∀ x $\in$ [0, 3] is __

x

f (x) = (1 - 2 x)^{2}

f (x) = 2 x^{2} + 4 x - 16

f (x) = - x^{2} - 6 x - 8

Join BYJU'S Learning Program