Question

Calculus Help
*MEDAL*

Answer 1

To find the relative extrema, you take the first derivative and find the critical points by setting the first derivative to zero and solve for x.

To find the inlfection point, you take the second derivative and find the critical points by setting the second derivative to zero and solve for x

Answer 2

Okay. for that I got x=0 and -0.5

Answer 3

I am just having problems getting the derivatives of the equation, since its so complicated

Answer 4

For the relative extrema, after finding the critical points, you have to see what part of the graph is increasing or decreasing. Same thing goes with finding the inflection point, there has to be a change in concavity.

Answer 5

how would i do that?

Answer 6

To find the inflection point, you have to first take the second derivative, set it to zero, get the critical point, do the second derivative test, and see which critical point changed in concavity.

Answer 7

That critical point that changed concavity is the inflection point.

Answer 8

i get that, its the getting the fricative of the equation

Answer 9

\(\color{blue}{Getting~The~Relative~Extrema}\)

\(f(x)=9x^{1/3}+9.2x^{4/3}\)

\(f'(x)=\Large \frac{46x^{4/3}}{5}+9x^{1/3}\)

set the derivative to zero

\(\Large \frac{46x^{4/3}}{5}+9x^{1/3}=0\)

\(\Large \frac{1}{5}x^{1/3}(46x+45)=0\)

\(x=0 ~and~x=\Large -\frac{45}{46}\)

Answer 10

its 9/2 not 9.2 but thank you. I got 0 and -0.5/ Plugging those into the original equation i got 0 and -5.36.

Answer 11

Yeh, i meant to write 9/2 but i made a typo mistake

Answer 12

nayways, so you -5.36 be the point?

Answer 13

-45/46 is approximately -0.98