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For what value(s) of k does the graph of g(x) have a normal line whose slope is -1/5 when x = 1?
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\[g(x)=k e^{^{2x}} + 3x\]
You can do this without help, right?
well I think I can. I need someone is double check my process and solution
I took the derivative of g(x), then I set it equal to -1/5 and sub x=1
I think k is \[k=\frac{ -8 }{ 3e^2 }\]
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I am using what you told me on the previous problem
Derivative gives you slope of tangent, not slope of normal.
so I need to use 5 instead of -1/5
yes
ok give me a moment to work it out
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k=1/(e^2)
am I correct?
yes
thanks I appreciate all of your help
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