Sketching the derivative or a function (1 Viewer)

[bemer]

This topic is hard.
When differentiating:
- the x intercepts turn into ???
- the turning points into??
- the point of inflexion into?

The opposite is integrating what do u do there

Any one noe

 

[Timothy.Siu]

ok, if u have a function and u have to sketch the derivative.....

the turning/stationary point of the function becomes zero on the derivative
if the function is increasing, derivative is POSITIVE (doesn't have to be increasing)
similarly for decreasing.

 

[tku336]

and the point of inflection of the function becomes a max/min on the derivative sketch and a zero on the second derivative sketch (I think)

 

[jet]

and the point of inflection of the function becomes a max/min on the derivative sketch and a zero on the second derivative sketch (I think)

Correct.

You don't really worry about x-intercepts. If you want an example ask, and I will post up a graph.

 

[bemer]

Correct.

You don't really worry about x-intercepts. If you want an example ask, and I will post up a graph.

YES plzz
ur a legend

 

[jet]

Okay, so I graphed f(x) = x³ - 6x² + 3x + 10 and the derivative f'(x) = 3x² - 12x + 3.

You can see that the turning points of the original function become the x-intercepts of the derivative. You should draw dotted lines from the T.P's of the original to the x-axis to show this.

You can see that the Point of inflexion on the original turns into the turning point of the derivative.

Wherever the original is increasing, (i.e the derivative is +ve) then the derivative graph is above the x-axis. The opposite occurs for a decreasing section of the function.

Finally, nothing happens with the x-intercepts of the original.
Hope it helps:

 

[bemer]

k thanks
do u do the same if ur integrating it or is their no such thing

 

[jet]

You don't need to worry about integrating unless you're doing 4 unit.