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aksik [14]
3 years ago
10

Geometry math question

Mathematics
1 answer:
kobusy [5.1K]3 years ago
8 0

the answer that you are looking for is d

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F(x)=(x-6)e^-3x find the interval in which f(x) is increasing, the intervals on which f(x) is decreasing, and the local extrema
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\bf f(x)=(x-6)e^{-3x}\\\\
-----------------------------\\\\
\cfrac{dy}{dx}=1\cdot e^{-3x}+(x-6)-3e^{-3x}\implies \cfrac{dy}{dx}=e^{-3x}[1-3(x-6)]
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\cfrac{dy}{dx}=e^{-3x}(19-3x)\implies \cfrac{dy}{dx}=\cfrac{19-3x}{e^{3x}}

set the derivative to 0, solve for "x" to get any critical points
keep in mind, setting the denominator to 0, also gives us critical points, however, in this case, the denominator will never be 0, so... no critical points from there

there's only 1 critical point anyway, and do a first-derivative test on it, check a number before it and after it, to see what sign the derivative has, and thus, whether the graph is going up or down, to check for any extrema
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Solve the system be either substitution or elimination 3x-5y=21 and 2x+y=1
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I used substitution and rearranged the second equation to y=-2x+1. Then I replaced the y in the first equation with -2x+1. So, I got 3x-5(-2x+1)=21. When you solve that you get 13x=26 or x=2. Then you replace x with 2 in either equation and end up with y=-3.

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