Find the largest and smallest values ...
- Thread starter al-horia
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Find the largest and smallest values of the given function
%20=%20t\tfrac{3}{2}%20e%20^{-2t}%20for%200%20\leq%20t%20\leq%201)
%20=%20(\frac{3}{2}%20t\tfrac{1}{3})%20(e^{-2t})%20+%20(t\tfrac{3}{2})%20(e^{-2t})%20(-2))
%20=%20(\frac{3}{2}%20t\tfrac{1}{3}%20+%20-2%20t\tfrac{3}{2})%20(e^{-2t}))
= 0%20=%20(e^{-2t})%20(t\tfrac{1}{2})%20(\frac{3}{2}-2t))
when t = 0 or t = 3/4
g(0)= 0
g(1) = .135
g(3/4)= .145
smallest value (0,0)
largest value ( 3/4 , .145 )
Is my work correct ?
I think in your second and third steps you have a typo. It should be t^(1/2), not t^(1/3), but your fourth step suddenly has the correct term of t^(1/2). Other than that, looks good to me.
al-horia,
Pay attention to srmichael's comment about the typo.
Your work is fine, good job!!
You have found where the maxima and minima occur by using g'(x). In order to know which value of x gives the maxima and which one gives the minima, there is another standard procedure.
What you have done is perfectly fine, but if it calls for it,
You should also find g''(x). If g''(a)<0, then at x=a you have a maxima, and if g''(a)>0, then at x=a you have a minima. You can check it with your example.
Here, a is the root of the equation g'(x)=0.
Cheers,
Sai.
Pay attention to srmichael's comment about the typo.
Your work is fine, good job!!
You have found where the maxima and minima occur by using g'(x). In order to know which value of x gives the maxima and which one gives the minima, there is another standard procedure.
What you have done is perfectly fine, but if it calls for it,
You should also find g''(x). If g''(a)<0, then at x=a you have a maxima, and if g''(a)>0, then at x=a you have a minima. You can check it with your example.
Here, a is the root of the equation g'(x)=0.
Cheers,
Sai.
al-horia,
when you answer the largest and/or the smallest values
of the function, you are to give the function values, not the co-ordinates.
Find the largest and smallest values of the given function
= 0
If you're going to use Latex (or Latex-like) characters, make use of the "^" for
exponentiation (or an appropriate different command) that achieves the exponentiation:
\(\displaystyle g(t) \ = \ t^{\frac{3}{2}}e^{-2t} \ \ \ for \ \ 0 \le t \le 1\)
\(\displaystyle g'(t) \ = \ \bigg(\dfrac{3}{2}\bigg)(t^{\frac{1}{3}})(e^{-2t}) \ + \ (t^{\frac{3}{2}})(e^{-2t})(-2)\)
And so on.
Is my work correct ?
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