Question

help please 12.

http://prnt.sc/ayllmn

Answer 1

\[f(x) = 7+6x^\frac{ 1 }{ 3 }\]
\[f'(x) = \frac{ 1 }{ 3 } . 6x ^{\frac{ -2 }{ 3 }} = 2x^\frac{ -2 }{ 3 }\]
\[f''(x) = \frac{ -4 }{ 3 } x^\frac{ -5 }{ 3 }\]

The inflexion is easiest to find. It will be a vertical inflexion when f'(x) is undefined (when x = 0). Concavity is described by the second derivative f''(x). f''(x) is always negative for all positive x and positive for all negative values of x(remember this is an odd powered root so we can take negative values of x). Thus concave up for x < 0, vertical inflexion at x = 0 and concave down for x >0. Note: A horizontal inflexion occurs for f'(x) = 0 and our first derivative is always positive given the power of x.

Answer 2

forget about the last line after the part I mention the horizontal inflexion occurs for f'(x) = 0, that's an typo.

Answer 3

can the inflection poiint can be solved algebraically ?

Answer 4

@xGuardians

Answer 5

Is a calculator allowed on this portion :)?

Answer 6

yes lol

Answer 7

That's like the most important factor because it makes things 100x easier haha.

Answer 8

oh haha

Answer 9

quetsion , once i found that my c= 0 doesnt that point get pluggin into the original question ?

Answer 10

So the graph generally looks something like this:
https://gyazo.com/6155bc0a9dd6149218fe1d90031f54bf
Now we can use the graph of the derivative and second derivative that @mww provided us with to help us understand the problem better.

Answer 11

oh but wouldnt the inflection pointbe f(0)= 7 ?

Answer 12

So to find where it is concave up or down, we have to input points into the second derivative.
The graph of y = f (x) is concave upward on those intervals where y = f "(x) > 0
.The graph of y = f (x) is concave downward on those intervals where y = f "(x) < 0.
If the graph of y = f (x) has a point of inflection then y = f "(x) = 0.

Answer 13

oh but im sort of confuse about the inflection though

Answer 14

To find the point of inflection set the second derivative equal to 0.
\[0 = \frac{ -4 }{ 3 } x^\frac{ -5 }{ 3 } \]

We can already see, when x = 0, there is a point of inflection.

Answer 15

so then can the x ^.. go on the bottom ?

Answer 16

Here, I'll help you understand a graph.
Take for instance the first graph we were presented with.
When you evaluate the slope at EVERY point. You would realize that this entire graph has a positive slope. Therefore, the first derivative will always > 0.

Answer 17

https://gyazo.com/b5ca5b9c93bd437bba3a07223900c94f

Answer 18

Then, when you look a look at the derivative of the slope and wonder as the slope increases, does it increase or decrease as it goes up. Take a look at the graph.

Answer 19

https://i.gyazo.com/2f90057e69a954aca916bb5c74e21bcd.png

Answer 20

A point of inflection will occur when it changes concavity.

Answer 21

oh okay so there would be no inflection point ? the graph keeps increasing

Answer 22

how would i find the cd / cu ?

Answer 23

Yes, at y = 0 there is one, since it changes concavity there.

Answer 24

I'm not sure what you mean by cd / cu

Answer 25

concave up and concave down

Answer 26

Refer to the second graph that I posted. From -infinity to 0, the slope of the graph is increasing until it becomes a vertical line, which is defined as concave up. Afterwards, from 0 to positive infinity, the slope of the line keeps decreasing, until it will eventually become horizontal.

Answer 27

https://i.gyazo.com/2f90057e69a954aca916bb5c74e21bcd.png

Answer 28

oh okay but can you help me solve it algebraically to find the concave up and down

Answer 29

\[f''(x) = \frac{ -4 }{ 3 } x^\frac{ -5 }{ 3 }\]
Take the second deriviative that you found earlier and plug in test x-values to see if they are positive or negative. For example, at x = 1.
\[0 = \frac{ -4 }{ 3 } (1)^\frac{ -5 }{ 3 } = -4/3\]
Since the answer is negative, it will be concave down at x = 1 and therefore from 0 to infinity.
\[0 = \frac{ -4 }{ 3 } (-1)^\frac{ -5 }{ 3 } = 4/3\]
Since the answer is positive, it will be concave down at x = -1 and therefore from 0 to negative infinity.
\[0 = \frac{ -4 }{ 3 } (0)^\frac{ -5 }{ 3 } = 0\]
Since the answer is zero, there will be a point of inflection at x = 0.
Reasoning:
The graph of y = f (x) is concave upward on those intervals where y = f "(x) > 0.
The graph of y = f (x) is concave downward on those intervals where y = f "(x) < 0.
If the graph of y = f (x) has a point of inflection then y = f "(x) = 0.