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f(x)=3(x)^(1/2)e^-x 1.Find the interval on which f is increasing 2.Find the interval on which f is decreasing 3.Find the local maximum value of f 4.Find the inflection point 5.Find the interval on which f is concave up 6.Find the interval on which f is concave down

Anyone can explain? I know the f'(x)=e^-x(3-6x)/2(x)^(1/2)

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up vote 2 down vote accepted




$$f'(x) = \dfrac{3 e^{-x}}{2 \sqrt{x}} - 3 e^{-x} \sqrt{x}$$

$$f''(x) = \dfrac{3 e^{-x}}{4 x^{3/2}} - \dfrac{3 e^{-x}}{\sqrt{x}} + 3 e^{-x} \sqrt{x}$$

  • A plot shows:

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  • $(1.)$ If $f'(x) > 0$ on an interval I, then $f$ is increasing on that interval.
  • $(2.)$ If $f'(x) < 0$ on an interval I, then $f$ is decreasing on that interval.
  • $(3.)$ Find values of $x$ where $f'(x)=0$ or $f'(x)$ is undefined. These will be candidates for possible maximum or minimum. Test these points further to determine which ones correspond to a maximum, a minimum or neither.
  • $(4.)$ If $f$ has a point of inflection at $c$, then either $f''(c)=0$ or $f''(c)$ is undefined.
  • $(5.)$ The sign of $f''(x)$ tells us if $f$ is concave up or down. More specifically, if $f''(x) > 0$ on an interval I, then $f$ is concave up on that interval.
  • $(6.)$ If $f''(x) < 0$ on an interval I, then $f$ is concave down on that interval.
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Another UV to go with the green! – amWhy Nov 16 '13 at 1:46


i) To find the interval of increasing, solve $f'(x)>0$ and for decreasing case, solve $f'(x)<0$.

ii) For local maximum, solve $f'(x)=0$ and do the second derivative test.

iii) for inflection points solve $f''(x)=0 $.

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Stepwise solutions $\cong +1$ – Babak S. Nov 15 '13 at 12:18
@B.S.: You are right. Thanks for the comment. – Mhenni Benghorbal Nov 15 '13 at 19:53

You have the first derivative of your function; so you are able to already answer the first three questions. Remembering that Exp[x] is always positive, then the sign of the derivative is the sign of (3-6x) [you could have simplified this]. Since x must be positive in order the function being defined in R (just because of the term Sqrt[x]), the derivative is positive for 0 < x < 1/2, zero for x = 1/2 and negative for any x > 1/2. I suppose you can continue with this, taking into account all the goods answers given to you while I was typing. If this is not clear, contact me.

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