Question

Se considera functia f:R⇒R, f(x)=x²+2x+m. Determinati m∈R pt care minimul functiei f este egal cu 2.

Answer 1

f'(x) = 2x+2    ⇒ 2x + 2 = 0      x = - 1
f(-1) = 1 - 2 + m = 2    m = 3

Answer 2

$f( x_{min} )= \frac{-b^2-4ac}{4a}$
-Δ=-4+4m
$2= \frac{-4+4m}{4}$
8=-4+4m
12=4m
m=3