1st, 2cd derivative plus key points of functions.

ffuh205

Junior Member
Joined
Mar 17, 2010
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51
Ok. This is what I have. Graph and describe (list Domain, Range, Local and Global Extrema = maximum and Minimum, points of inflection, Intervals of Increase, Decrease, Curving up or down). Graph the function and its first and second derivatives in different colors on the same axes and label them. F(x) =x^2•LN(x^4)•e^(- 2•x). Ok. for the first derivative I have 2x, 4x^3/x, & 2e^(2x) for each section but I'm not sure how to find a triple product. Beyond that, I am lost.
 
\(\displaystyle Well \ do \ it. \ We're \ not \ here \ to \ spoon \ feed \ you, \ besides \ I'm \ out \ of \ Gerbers.\)
 
\(\displaystyle F(x) \ = \ x^{2}[ln(x^{4})]e^{-2x}\)

\(\displaystyle F'(x) \ = \ 2x[ln(x^{4})]e^{-2x}+x^{2}\bigg(\frac{4x^{3}}{x^{4}}\bigg)e^{-2x}+x^{2}[ln(x^{4})]e^{-2x}(-2), \ now \ simplify.\)
 
Ok. So that would be F’(x) = x*e^-2x [x*ln(x^4) + 4 -2x*ln(4^x)].

F"(x) would then be [2ln(x^4)*e^-2x] + (8*e^-2x) + [2xln(x^4)*e^-2x] + (4e^-2x) +(4xe^-2x) + [2xln(x^4)*e^-2x] + (4xe^-2x) - [2x^2ln(x^4)*e^-2x]. You could factor out the 2 which would give 2{[ln(x^4)*e^-2x] + (4*e^-2x) + [xln(x^4)*e^-2x] + (2e^-2x) +(2xe^-2x) + [xln(x^4)*e^-2x] + (2xe^-2x) - [x^2ln(x^4)*e^-2x] }
 
\(\displaystyle F'(x) \ = \ 2x[ln(x^{4})]e^{-2x} \ + \ 4xe^{-2x}-2x^{2}[ln(x^{4})]e^{-2x}\)

\(\displaystyle F'(x) \ = \ 2xe^{-2x}[ln(x^{4})+2-xln(x^{4})], \ that \ should \ do \ it.\)

\(\displaystyle However, \ F'(x) \ = \ 2xe^{-2x}[2+ln(x^{4})-ln(x^{4x})] \ = \ 2xe^{-2x}\bigg[2+ln\bigg(\frac{x^{4}}{x^{4x}}\bigg)\bigg]\)

\(\displaystyle = \ 2xe^{-2x}[2+ln(x^{4-4x})] \ = \ 2xe^{-2x}[2+[(4-4x)ln(x)]] \ = \ 2xe^{-2x}[2(1+[(2-2x)ln(x)])]\)

\(\displaystyle = \ 4xe^{-2x}[1+[(2-2x)ln(x)]]\)
 
\(\displaystyle I \ have \ no \ idea, \ if \ I \ was \ to \ find \ f"(x) \ for \ the \ above, \ I \ would \ avoid \ all \ the\)

\(\displaystyle grunt \ work \ and \ employ \ my \ Trusty \ TI-89.\)

\(\displaystyle It \ is \ 2nd \ not \ 2cd.\)
 
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