Question
Asked Dec 3, 2019
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5.) y=500/[1 + 24e(-.26t)] with an inflection point at t=12.2

Sketch the graph and its inflection point

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Expert Answer

Step 1

We find y\\' using exponent rule and chain rule.

 

500
у-.
1+24e0.26
y500(1+24e0.26)
-0.26)
-2
(1+24e0.26)
dt
y' 500(-1)+24e
y'500(1+24c26) (24e 0.26)(-0.26)
y' 3120e0.26(1+24e ~0.26)
-2
3120
у'3
0.26(1+24e 0.26)
3120e0.26
у-.
(e026+24)
>0
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Image Transcriptionclose

500 у-. 1+24e0.26 y500(1+24e0.26) -0.26) -2 (1+24e0.26) dt y' 500(-1)+24e y'500(1+24c26) (24e 0.26)(-0.26) y' 3120e0.26(1+24e ~0.26) -2 3120 у'3 0.26(1+24e 0.26) 3120e0.26 у-. (e026+24) >0

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Step 2

For all t , y'>0

So, y is an increasing function. 

Step 3

Use quotient rule to find the second ...

3120e
у"%3
0.26"(0.26) (е0.261.
"+24)*-3120e 0.262(e0.26t
+24)(0.26e0.2)
4
(e0.2t+24)
+24)-2e026]
3120e0.26t0.26t_
у"-
0.26t
+24) (0.26) [(e'
0.261 24)*
(e
0.20(e
0.26t
811.2e
у"—
+24) [24-e0.26t]
0.26
(e
+24)
help_outline

Image Transcriptionclose

3120e у"%3 0.26"(0.26) (е0.261. "+24)*-3120e 0.262(e0.26t +24)(0.26e0.2) 4 (e0.2t+24) +24)-2e026] 3120e0.26t0.26t_ у"- 0.26t +24) (0.26) [(e' 0.261 24)* (e 0.20(e 0.26t 811.2e у"— +24) [24-e0.26t] 0.26 (e +24)

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