critical points and inflection points

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lauren52

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Oct 28, 2008
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I need help in graphing, plotting critical points and inflection points of the function f(x) =x^4+8x^3+12
Thus far, I came up with the following; however, I am unsure if this is correct.
f'(x)=4x^3+24^2
4x^2(x+6)=0
4x^2=0
x+6=0 x=-6
critical point is -6 and 0

Plugging -6 into the original equation
-6^4+8(-6)^3+12
1296-1728+12=-420
and
Plugging in 0 into the original equation
0^4+8(0)^3+12=12

no relative minimum
relative maximum (-6,-420)

using the 2nd derivative to find inflection points
f''(x)=12x^2+48x
12x(x+4)=0
12x=0, x=-4

plugging 0 into the orginal equation
f(0)=0^4+8(0)^3+12=12
Inflection points (0,12)

f(-4)=4^4+8(4)^3+12=12
Inflection points (-4,12)

Any assistance you can provide would be greatly appreciated.

Thank you,
 
lauren52 said:
I need help in graphing, plotting critical points and inflection points of the function f(x) =x^4+8x^3+12
Thus far, I came up with the following; however, I am unsure if this is correct.
f'(x)=4x^3+24^2
4x^2(x+6)=0
4x^2=0
x+6=0 x=-6
critical point is -6 and 0

to decide on whether maxima or minima or point of inflection you need to use second derivative test.

Plugging -6 into the original equation
-6^4+8(-6)^3+12
1296-1728+12=-420
and
Plugging in 0 into the original equation
0^4+8(0)^3+12=12

no relative minimum
relative maximum (-6,-420)

using the 2nd derivative to find inflection points
f''(x)=12x^2+48x
12x(x+4)=0
12x=0, x=-4

plugging 0 into the orginal equation
f(0)=0^4+8(0)^3+12=12
Inflection points (0,12)

f(-4)=4^4+8(4)^3+12=12
Inflection points (-4,12)

Any assistance you can provide would be greatly appreciated.

Thank you,
Click to expand...
 
Hello, lauren52!

You did pretty good . . . and thank you for showing your work!

I need help in graphing, plotting critical points and inflection points

. . of the function .\(\displaystyle f(x) \:=\:x^4+8x^3+12\)

Thus far, I came up with the following:
. . \(\displaystyle f'(x)\:=\:4x^3+24^2\:=\:0 \quad\Rightarrow\quad 4x^2(x+6)\:=\:0\)

. . \(\displaystyle \begin{array}{cccccc}4x^2\:=\:0 & \Rightarrow & x \:=\:0 \\ x+6\:=\:0 & \Rightarrow & x\:=\:\text{-}6 \end{array}\)

Critical points: .\(\displaystyle (0,12),\;(\text{-}6,\text{-}420)\) . . . . Good!

no relative minimum . . . . no
relative maximum (-6,-420) . . . . no
Click to expand...

Use the Second Derivative Test to check out the critical values.

\(\displaystyle f''(x) \:=\:12x^2 + 48x \:=\:12x(x+4)\)

\(\displaystyle \text{At }x=0\!:\;\;f''(0) \:=\:12(0)(4) \:=\:0\quad\hdots\quad \text{Could be an inflection point}\)

\(\displaystyle \text{At }x = \text{-}6\!:\;\;f''(\text{-}6) \:=\:12(\text{-}6)(\text{-}2) \:=\:\boxed{+}144 \quad\hdots \quad\text{positive: concave up }\;\cup\;\text{ ... minimum at }(\text{-}6,\text{-}420)\)


For inflection points, solve \(\displaystyle f''(x) \:=\:0\)

. . \(\displaystyle 12x(x+4) \:=\:0 \quad\Rightarrow\quad x \:=\:0,\text{-}4\)

\(\displaystyle \text{Inflection points: }\:(0,12),\;(\text{-}4,\text{-}244)\) .**


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

** .It is advisable to test these inflection points, too.

Recall that an inflection point is where the concavity changes.


We suspect that \(\displaystyle x = 0\) is an inflection point.
Evaluate \(\displaystyle f''(x) \:=\:12x(x+4)\) on "both sides".

. . \(\displaystyle \begin{array}{ccccccc}f''(\text{-}1) \:=\:(\text{-}12)(3) \:=\:\boxed{\text{-}}\,36 & \text{negative: concave down} & \cap \\ \\[-3mm] f''(1) \:=\:(12)(5) \:=\:\boxed{+}\,60 & \text{positive: concave up} & \cup \end{array}\)

To the left of \(\displaystyle x = 0\), the graph is concave down.
To the right of \(\displaystyle x=0\), the graph is concave up.
. . The concavity does change at \(\displaystyle x = 0.\)

Therefore: .(0,12) is an inflection point.

. . . Get the idea?

 
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