The question is Find c such that `f'(c)=(f(3)-f(0))/3=(e^-3-1)/3`

`f'(x)=-3(e^(-3x))`

If `f'(c)=(e^(-3)-1)/3, `

then `-3(e^(-3c))=(e^(-3)-1)/3, `

Let's solve for c,

`e^(-3c)=-(e^(-3)-1)/9=(1-e^(-3))/9`

`ln(e^(-3c))=ln((1-e^(-3)/9)`

`-3c=ln((1-e^(-3)/9)`

**Answer** c=-1/3 ln((1-e^(-3)/9) **or equivalently** ` ` `lc=1/3 n(9/(1-e^(-3)))~~ 0.75`

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