Question

I need to find the intervals on which the following equations have zeroes, extrema, points of inflection and intervals increasing or decreasing, as well as concave up or down. I need to show these on a graph but is there a better way to present the data (before graphing) instead of just x=1 and sub into equation, then x=2, and so on? Thanks in advance

Answer 1

Equation 1) \[f(x) =x^2 +\frac{ 1 }{ x }\]
Equation 2) \[h(x) =x^3-9x^2+27x\]

Answer 2

Do you know how to take the derivative?

Answer 3

yes

Answer 4

K what is the derivative of the first one?

Answer 5

2x -1x^-2 ?

Answer 6

right, now set that =0 and solve for x

Answer 7

not sure how to do that for the -x^-2 part? Do you add 1^2 to RHS?

Answer 8

to give\[x=\frac{ 1 }{ 2 }\]

Answer 9

woops sorry

Answer 10

\(2x-\frac{1}{x^2}=0\implies 2x^3-1=0\implies x=\frac{1}{\sqrt[3]{2}}\)

Answer 11

ok

Answer 12

so this gives the zero value.

Answer 13

Ok, so we have some points to look at.

First we have \(0\) will give an asymptote so we look to the left of \(0\)

We can put in any number to the left of \(0\) in the derivative and get a negative number so we know the original equation is decreasing on the left of \(0\).

So now we at the derivative between \(0\) and \(\frac{1}{\sqrt[3]{2}}\) and get that it is positive so the original equation is also decreasin on \((0, \frac{1}{\sqrt[3]{2}})\).

Do the same thing to the right of the fraction solution to get its decreasing.

you with me?

Answer 14

Finally we need to look at the zeros of the original equation and the right and left limits at 0.

Answer 15

Tie all the together and you get the graph.

Answer 16

I am with you

Answer 17

Also plug in \(\frac{1}{\sqrt[3]{2}}\) to see that it is the min for \(x>0\)

Answer 18

Ok, and we tell that the function is decreasing by comparing the values of each x=number that we are using?

Answer 19

we get