Answer to Question #314790 in Real Analysis for Jyo

Question

Find the relative extrema for

i) 𝑓( 𝑥 )= 𝑥^3 − 3𝑥 + 5

𝑖𝑖) f(x)=𝑥^4 + 2𝑥^2 − 4

Answer 1

i) $f(x)=x^3-3x+5$

$f’(x)=3x^2-3=0$

$x=\pm1$ ;

$f’(-2)=12-3>0$

$f’(0)=-3<0$

$f’(2)=12-3>0$

Thus $x=-1$ - maximum; $x=1$ - minimum.

Answer: $x_{max}=-1$ ; $x_{min}=1$ .

ii) $f(x)=x^4+2x^2-4$

$f’(x)=4x^3+4x=4x(x^2+1)=0$

$x=0$

$f’(-1)=-8<0$

$f’(1)=8>0$

$x=0$ - minimum.

Answer: $x_{min}=0$ .