# How to find inflection points? ?

Given f(x)=0.5x^(4)+2x^(3)-9x^(2)+12

I got the answer of -3,82.5 is that correct?

### 3 Answers

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• 1 month ago
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Find when the 2nd derivative is equal to 0

f(x) = 0.5 * x^4 + 2 * x^3 - 9 * x^2 + 12

f'(x) = 2 * x^3 + 6 * x^2 - 18 * x

f''(x) = 6x^2 + 12x - 18

f''(x) = 0

0 = 6x^2 + 12x - 18

0 = x^2 + 2x - 3

3 = x^2 + 2x

4 = x^2 + 2x + 1

4 = (x + 1)^2

x + 1 = -2 , 2

x = -3 , 1

There are 2 inflection points

f(-3) = 0.5 * 3^4 + 2 * 3^3 - 9 * 3^2 + 12

f(-3) = 0.5 * 81 + 2 * 27 - 9 * 9 + 12

f(-3) = 40.5 + 54 - 81 + 12

f(-3) = 94.5 + 12 - 81

f(-3) = 13.5 + 12

f(-3) = 25.5

f(1) = 0.5 * 1^4 + 2 * 1^3 - 9 * 1^2 + 12

f(1) = 0.5 + 2 - 9 + 12

f(1) = 14.5 - 9

f(1) = 5.5

(-3 , 25.5) , (1 , 5.5)

• Philip
Lv 6
1 month ago

f(x) = 0.5x^4 +2x^3 -9x^2 +12, f'(x) = 2x^3 +6x^2 -18x, f''(x) = 6(x^2 +2x -3). Points of

inflection occur where f''(x) = 0, ie., x^2+2x-3 = 0, ie., (x+3)(x-1) = 0, ie., x = -3...(i) or

x = 1...(ii). For (i) holding, f(-3) = (1/2)81 -2*27 -81 + 12 = -42 -81/2 = -165/2 = -82.5.

Then one point of inflection = (-3 , -82.5). For (ii) holding, f(1) = (1/2)+2-9+12 = 5.5.

Then 2nd point of inflection = (1 , 5.5).

• 1 month ago

Points of inflection occur where the second derivative changes signs. According to the Intermediate Value Theorem, the second derivative can only change sign if it is discontinuous or if it passes through zero, so let's take the second derivative and set it equal to zero.

f'(x) = 2x^3 + 6x^2 - 18x

f''(x) = 6x^2 + 12x - 18 = 0

Divide both sides by 6:

x^2 + 2x - 3 = 0

(x - 1)(x + 3) = 0

x = 1 or -3

Here, both of the roots have multiplicity one, so the second derivative changes sign at both roots. This means they will both be points of inflection.

Then, for the y-values, just plug the x-values into the original function. (not the derivatives!)

f(1) = 5.5, f(-3) = -82.5

So the points of inflection will occur at both (-3, -82.5) and (1, 5.5).

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