# Critical Points, Inflection Points, Increasing/Decreasing Functions

• Oct 29th 2009, 07:38 PM
iheartthemusic29
Critical Points, Inflection Points, Increasing/Decreasing Functions
1). f(x)=ln(2x^2+1)
Use derivatives to find the x-values of any critical points and inflection points exactly.
critical points (enter as a comma-separated list): x= 0
inflection points (enter as a comma-separated list): x= ???

The critical point is zero, but I'm not sure what the inflection point is. I thought it was also zero, but that is incorrect.

2). For the function f(x)=2x^3−6x+7, find all intervals where the function is increasing.
Similarly, find all intervals where the function is decreasing.
I know you take the derivative, which is 6x^2-6. I believe then you set it equal to zero and then test points. However, I'm not getting the correct answer.

3). For the function f(x)=8x+5sin(x). The derivative is 5cosx+8. Find all intervals where the function is increasing.
Similarly, find all intervals where the function is decreasing.

• Oct 29th 2009, 07:58 PM
ux0
Quote:

Originally Posted by iheartthemusic29
1). f(x)=ln(2x^2+1)
Use derivatives to find the x-values of any critical points and inflection points exactly.
critical points (enter as a comma-separated list): x= 0
inflection points (enter as a comma-separated list): x= ???

The critical point is zero, but I'm not sure what the inflection point is. I thought it was also zero, but that is incorrect.

a point on a curve at which the second derivative changes sign.

therefore we have...

$\displaystyle f'(x)=\frac{4x}{2x^2+1}$
$\displaystyle f''(x)=\frac{-8x^2+4}{(2x^2+1)^2}$

So if u solve the numerator by setting it equal to zero you get two points..

$\displaystyle x=\sqrt{\frac{1}{2}}$

$\displaystyle x=-\sqrt{\frac{1}{2}}$

Now check a number smaller then $\displaystyle x=-\sqrt{\frac{1}{2}}$

a number in between them

and a number larger then $\displaystyle x=\sqrt{\frac{1}{2}}$

If the signs change at theses points, then its an inflection.
• Oct 29th 2009, 08:02 PM
tonio
Quote:

Originally Posted by iheartthemusic29
1). f(x)=ln(2x^2+1)
Use derivatives to find the x-values of any critical points and inflection points exactly.
critical points (enter as a comma-separated list): x= 0
inflection points (enter as a comma-separated list): x= ???

The critical point is zero, but I'm not sure what the inflection point is. I thought it was also zero, but that is incorrect.

Indeed the only critical point is x=0 and since f'(x) changes sign when passing through x= 0 (from minus to plus) this is a minimum point.
Since $\displaystyle f''(x)=-4\frac{(x-1)(x+1)}{(2x^2+1)^2}$, check where this 2nd derivative vanishes for POSSIBLE inflexion points, and then check which derivative after the 2nd one is the first one that does NOT vanish at some of these points: if these last derivative is an odd one then you have an inflexion point, othewise you haven't.

2). For the function f(x)=2x^3−6x+7, find all intervals where the function is increasing.

If a function is derivable over some interval, then it is increasing in this interval iff f'(x) > 0 in the interval. From here deduce where f increases and where it decreases.

Similarly, find all intervals where the function is decreasing.
I know you take the derivative, which is 6x^2-6. I believe then you set it equal to zero and then test points. However, I'm not getting the correct answer.

3). For the function f(x)=8x+5sin(x). The derivative is 5cosx+8. Find all intervals where the function is increasing.

Do as in (2)...you'll get a very nice and, perhaps, slighty unexpected answer

Tonio

Similarly, find all intervals where the function is decreasing.