Find Min/Max/Inflection Point for f(x)=3x^5-10x^3+1

[Harmony]

Homework Statement

The function f is defined as
f(x)=3x^5-10x^3+1
1) Determine the maximum and minimum points, as well the point of inflection of the graph f.

Homework Equations

All the differentiation equation.


3. The Attempt at a Solution
I found the maximum point and minimum point, but I had some trouble with the inflection point.

So...
f''(x)=60x^3-60x^2
f''(x)=0,
60x^3-60x^2=0
x=0, x=1

f'''(0)=0

Hence the only inflection point is (1,-6)

But the answer given is (0,1),(1,-6)and (-1,8). Is the answer correct?

 

[jing]

Not sure this should be in this topic as it is calculus not precalculus.

Have another look at your differentiation, in particular check the term -60x^2 in your expression for f''(x).

 

[marlon]

f''(x)=60x^3-60x^2

This second derivative is wrong. Calculate it again.

Your approach is correct though.

marlon

 

[HallsofIvy]

Homework Statement

The function f is defined as
f(x)=3x^5-10x^3+1
1) Determine the maximum and minimum points, as well the point of inflection of the graph f.

Homework Equations

All the differentiation equation.


3. The Attempt at a Solution
I found the maximum point and minimum point, but I had some trouble with the inflection point.

So...
f''(x)=60x^3-60x^2
f''(x)=0,
60x^3-60x^2=0
x=0, x=1

f'''(0)=0

Hence the only inflection point is (1,-6)

But the answer given is (0,1),(1,-6)and (-1,8). Is the answer correct?

f"(x)= 60x³- 60x= 60x(x²- 1)= 0
for x= 0, 1, and -1.

One definition of "inflection point" is that the second derivative changes sign there. Yes, it is true that since f '''(0)= 0, the second derivative might NOT change signs there. For example, if f ''= x², then while f''(0)= 0, f'' does not change signs there. But f'''(0)= 0 does not mean it CAN'T change signs there. For example, if f ''= x³, then, again, f'''(x)= 3x² so f'''(0)= 0 but x³ does change signs there.

In this particular case, f''(1/2)= 60/8- 120< 0 while f''(-1/2)= -60/8+120> 0 so f'' does change signs at 0.