Question

Find the absolute maximum (value) of
f(x)=x^2+4x+1 on [-5, 0]

Answer 1

f(x) is a parabola that opens up so its vertex is a minimum

f'(x) = 2x+4 = 0
--> x = -2
Vertex at x = -2
Parabolas are symmetric, so f(-1) = f(-3) and f(0) = f(-4) and f(1)=f(-5) and f(x) is increasing the farther away from the vertex it gets
therefore maximum value in interval [-5,0] would have to be at x = -5

f(-5) = 6