The minimum value of f(x) = e(x4-x3+x2) is

Question

The minimum value of f(x) = e(x4-x3+x2) is (A) e (B) -e (C) 1 (D) -1

Tardigrade Admin · Accepted Answer

f(x)=e(x4-x3+x2), f'(x)=ex4-x3+x2 (4 x3-3 x2+2 x) ⇒ f(x) is decreasing for x < 0, increasing for x > 0. ∴ Minimum is at x=0 Hence, minimum value of f(x)=f(0)=e0=1

Q. The minimum value of $f (x) = e^{(x^{4} - x^{3} + x^{2})}$ is

$e$

$- e$

$1$

$- 1$

$f (x) = e^{(x^{4} - x^{3} + x^{2})}, f^{^{'}} (x) = e^{x^{4} - x^{3} + x^{2}}$
$(4 x^{3} - 3 x^{2} + 2 x)$
$\Rightarrow f (x)$ is decreasing for $x < 0$ ,
increasing for $x > 0$ .
$∴$ Minimum is at $x = 0$
Hence, minimum value of
$f (x) = f (0) = e^{0} = 1$

Q. The minimum value of f(x)=e(x4−x3+x2) is

e

−e

1

−1

f(x)=e(x4−x3+x2),f′(x)=ex4−x3+x2 (4x3−3x2+2x) ⇒f(x) is decreasing for x<0, increasing for x>0. ∴ Minimum is at x=0 Hence, minimum value of f(x)=f(0)=e0=1

Q. The minimum value of $f (x) = e^{(x^{4} - x^{3} + x^{2})}$ is

$e$

$- e$

$1$

$- 1$

$f (x) = e^{(x^{4} - x^{3} + x^{2})}, f^{^{'}} (x) = e^{x^{4} - x^{3} + x^{2}}$
$(4 x^{3} - 3 x^{2} + 2 x)$
$\Rightarrow f (x)$ is decreasing for $x < 0$ ,
increasing for $x > 0$ .
$∴$ Minimum is at $x = 0$
Hence, minimum value of
$f (x) = f (0) = e^{0} = 1$