Why is Zero not a Critical Point of f ( x ) = ( x 2 − 63 ) e − x ?

Question

Why is Zero not a Critical Point of   f  (  x  )  =  (      x    2    −  63  )      e          −      x      ?

CobeBedererKDs · Accepted Answer

The critical points are the points where the derivative is

0

. Since you have found the derivative to be

- e^{- x} (x - 9) (x + 7)

,

0

cannot be a critical point, because

- e^{- 0} (0 - 9) (0 + 7) = - 1 \cdot (- 9) \cdot 7 = 63 \neq 0

accessedg05 · Accepted Answer

The derivative is      f    ′    (  x  )  =  2  x      e          −      x        −  (      x    2    −  63  )      e          −      x        =  −      e          −      x        (      x    2    −  2  x  −  63  )  =  −      e          −      x        (  x  −  9  )  (  x  +  7  )A product is zero if and only if one of the factor is:      e          −      x        &amp;gt;  0  for all   x  x  −  9  =  0  if and only if   x  =  9  x  +  7  =  0  if and only if   x  =  −  7Thus only   −  7  and   9 are critical points.

Why is Zero not a Critical Point of f(x)=(x^2−63)e^(−x)?

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