whats the minimum of the function f(x)=-2x^2+4x+4

Question

i need help

Answer 1

take the derivativ & set it tu 0 tu get the lo or hi or maebee both

y=-2x^2+4x+4

dy/dx=-4x+4....=0=4*(1-x)

so it =0 at x=1

y=-2*1+4*1....4-2....y=2 at x=1

Need tu see if that be hi or lo, so tri y at x=0...

y=4 at x=0...bigger than y at x=1, so yu found a lo

Answer 2

f(x)=-2x²+4x+4 ··· Eq.1

The graph of a quadratic equation y=ax²+bx+c (a≠0) is a parabola.   If a>0 (y">0), the curve is concave downwards*, and has the minimum at its critical point (y'=0).   If a<0 (y"<0), the curve is convex upwards*, and has the maximum at its critical point (y'=0).

Therefore, the given equation doesn't have the minimum, but the maximum.

Here, we examine the maximum.   From Eq.1, we have:

f'(x)=-4x+4=-4(x-1), so at x=1, f'(1)=0

f"(x)=(f'(x))'=(-4x+4)'=-4 (<0)**

Therefore, the given function has the maximum at x=1 and the maximum is: f(1)=-2(1)²+4(1)+4=6

* f(x)=ax²+bx+c=ax²(1 + b/(ax) + c/(ax²)). If a>0, and x approaches (+) or (-) infinity, f(x) approaches (+) infinity: the graph is concave downwards.   If a<0, f(x) approaches (-) infinity: convex upwards.

** If f'(x)=0 (the tangent is parallel to x-axis), f(x)">0 at the minimum point.   If f'(x)=0, f(x)"<0 at the maximum point.