Question

Derivative Applications: Maximum and minimum through derivatives

Answer 1

9514 1404 393

Answer:

1. y'=10x+7; y''=10; vertex: (-0.7, 0.55), min; no real zeros
2. y'=-2x-5; y''=-2; vertex: (-2.5, 8.75), max; zeros: {-5.458, 0.458}

Step-by-step explanation:

For a general parabola ...

  y = ax² +bx +c

The first derivative is ...

  y' = 2ax +b

and the second derivative is ...

  y'' = 2a

The second derivative is a (non-zero) constant, so there is no point of inflection. The sign of the second derivative (the sign of 'a') tells you whether the graph opens upward (a>0) or downward (a<0).

The first derivative is 0 where ...

  y' = 0 = 2ax +b

  x = -b/(2a) . . . . . extreme point of the graph

Whether this point is a maximum (a<0) or a minimum (a>0) depends on the sign of 'a'.

The value of the function at the extreme is ...

  y = a(-b/2a)² +b(-b/2a) +c = b²/(4a) -b²/(2a) +c

  y = c -b²/(4a)

So, the extremum is ...

  (x, y) = (-b/(2a), c -b²/(4a)) . . . . . vertex of the parabola

__

1. We have a=5, b=7, c=3.

  First derivative: f'(x) = 10x +7

  Second derivative: f''(x) = 10 . . . . graph opens upward

  Zeros: the discriminant b² -4ac is 7²-4(5)(3) = -11, so no real zeros

  Vertex: (-7/10, 3 -49/20) = (-0.7, 0.55) . . . minimum

__

2. We have a=-1, b=-5, c=5/2.

  First derivative: f'(x) = -2x -5

  Second derivative: f''(x) = -2 . . . . graph opens downward

  Zeros: found from the quadratic formula: x = (-b±√(b²-4ac))/(2a)

     x = (-(-5) ±√((-5)² -4(-1)(5/2)))/(2(-1)) = (5 ±√35)/2 ≈ {-5.458, 0.458}

  Vertex: (-5/2, 5/2-25/-4) = (-5/2, 35/4) = (-2.5, 8.75) . . . maximum

Answer 2

Answer:

fx=5x

2

+7x+3

fx=−x

2

−5x+

2

5