Question

Help

Answer 1

a) Calculate Limits
b)relative extremes
c)inflection point

Answer 2

@JopHP

Answer 3

u want a general explain ?!
or to answer a specific question ?!

Answer 4

only answer

Answer 5

a) there is exponential properties we hav to remember it
That: limx→+∞ex=∞ & limx→−∞ex=0
Therefore: \[LimX-->\infty e^x (x^2-x+1) =\]
And
\[LimX-->\infty e^x (x^2-x+1)=0\]

Answer 6

The 1st =infinity

Answer 7

ok

Answer 8

the relative extrems are the local maximum & local minimum Right ?!

Answer 9

is the relative extrema of f, determine their maximum or minimum

Answer 10

Lol...Spanish...:)

Answer 11

yup

Answer 12

\[e^x(x^2-x+1)\]
Therefore: 1st derivative test
\[e^x(x^2-x+1) + (2x-1)e^x\]
by taking e^x common factor we get
\[e^x(x^2+x)\] Then we put this equation = to Zero
Therefore \[since e^x can't =0 \] So
\[x^2+x=0 .. x(x+1)=0\]
therefore \[x=0 & x=-1\]
therefore (0,0) & (-1,0) are the critical points then u put them on the number-line and substitute with no. before & no. after to get the local maximum & local minimum
& the point at which the curve changes from local maximum to local minimum is Called the inflection point

Answer 13

okk thanks

Answer 14

Wlcm

Answer 15

good work