Posted by drew on Monday, February 4, 2013 at 6:11pm.
1. y=x2 - 16
for maximum or minumum, dy/dx=0
hence 2x=0 or x=0
minimum value of function occurs when x=0 which is y=-16
2.using same principle as above,
dy/dx=10x+5
when dy/dx=0 => 10x+5=0 => x=-1/2
minimum value of y= -5/4 - 5/2 + 11
for 3, 4 and 5 use the same principle as 1 and 2
max: infinity (at x=inf)
min: x=1/2, y= you figure it. check sign.
max: infinity (at x=inf)
min: x=1/2, y= you figure it. check sign.
Not sure if you know Calculus, as done by Me
so let's complete the square
y = x^2 - x - 56
= x^2 - x + 1/4 - 1/4 - 56
= (x- 1/2)^2 - 225/4
so we have a min of -225/4 , when x = 1/2
(min because the parabola opens upwards)
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