Question

f(x)=x^4-5x^3
Curving the Sketch. Find the interval notations for the concavity when it concaves up and concaves down.

Answer 1

4x^3-15x^2
Critical values: 0,15/4
Increasing (15/4, infinity)
Decreasing (-infinity,0]U(0,15/4)
Local max: none
Local min: 15/4
I found all the way until there and I am stuck.

Answer 2

Find the second derivative.

Answer 3

Inflection points?

\[f'(x)=4x^{3}-15x^{2}\]

\[f''(x)=12x^{2}-30x^{1}=6x(2x-5)\]

Inflection points f''(x)=0 at x=0 and x=5/2

Answer 4

\[f(x)=x^4-5x^3 \]
\[f'(x)=4x^3-15x^2\]
\[f''(x)=12x^2-30x\]
At S.P's (Stationary Points) f'(x)=0
\[4x^3-15x^2=0\]
\[x^2(4x-15)=0\]
\[x=0\]
\[y=0\]
\[f''(x)=0\]
^^ that point MIGHT be an inflection. Don't take that for granted.
\[x=\frac{15}{4}\]
\[y=-\frac{16875}{256}\]
\[f''(x)>0\]
Therefore Minimum T.P (Turning Point) at (\frac{15}{4},-\frac{16875}{256})

Answer 5

\[(\frac{15}{4},-\frac{16875}{256})\]

Answer 6

So to find inflection points you would just take the second derivative and make it equal to zero?

Answer 7

Yeah.

Answer 8

Make sure you test the points beside the value so you know how the inflexion point looks like.

Answer 9

Thank you guys! I got it (:
Concave Up: (-Infinity,0)U(5/2,Infinity)
Concave down: (0,5/2)
Inflection points: x=0, 5/2

Answer 10

No worries.

Answer 11

As regards the point of inflection : you want the third derivative to be non-zero when the second derivative is zero.

For instance a line will have second derivative zero everywhere ... but the third derivative will also be zero.

Answer 12

Thank you for the clarification!