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Math  /  Calculus

QuestionMath 2211 Classwork over 4.1 and 4.3
1. (10 points) Consider f(x)=x4−6x2f(x)=x^{4}-6 x^{2}. (a) Find the critical numbers of the function. (b) Find the intervals on which ff is increasing or decreasing. (c) Find the local maximum and minimum values of ff. (d) Find the inflection points. (e) Find the intervals of concavity.

Studdy Solution

STEP 1

1. The function given is f(x)=x4−6x2 f(x) = x^4 - 6x^2 .
2. Critical numbers occur where the derivative is zero or undefined.
3. Increasing or decreasing intervals are determined by the sign of the first derivative.
4. Local maxima and minima occur at critical points where the first derivative changes sign.
5. Inflection points occur where the second derivative changes sign.
6. Intervals of concavity are determined by the sign of the second derivative.

STEP 2

1. Find the critical numbers of the function.
2. Determine the intervals on which f f is increasing or decreasing.
3. Find the local maximum and minimum values of f f .
4. Find the inflection points.
5. Determine the intervals of concavity.

STEP 3

To find the critical numbers, first find the derivative of f(x) f(x) :
f′(x)=ddx(x4−6x2)=4x3−12x f'(x) = \frac{d}{dx}(x^4 - 6x^2) = 4x^3 - 12x
Set the derivative equal to zero to find critical points:
4x3−12x=0 4x^3 - 12x = 0
Factor the equation:
4x(x2−3)=0 4x(x^2 - 3) = 0
Solve for x x :
4x=0⇒x=0 4x = 0 \quad \Rightarrow \quad x = 0
x2−3=0⇒x=±3 x^2 - 3 = 0 \quad \Rightarrow \quad x = \pm\sqrt{3}
The critical numbers are x=0,3,−3 x = 0, \sqrt{3}, -\sqrt{3} .

STEP 4

Determine the intervals on which f f is increasing or decreasing by testing intervals around the critical numbers in the first derivative:
Test intervals: (−∞,−3),(−3,0),(0,3),(3,∞) (-\infty, -\sqrt{3}), (-\sqrt{3}, 0), (0, \sqrt{3}), (\sqrt{3}, \infty) .
Choose test points: x=−2,−1,1,2 x = -2, -1, 1, 2 .
Evaluate f′(x) f'(x) at these points:
f′(−2)=4(−2)3−12(−2)=−32+24=−8(decreasing) f'(-2) = 4(-2)^3 - 12(-2) = -32 + 24 = -8 \quad (\text{decreasing})
f′(−1)=4(−1)3−12(−1)=−4+12=8(increasing) f'(-1) = 4(-1)^3 - 12(-1) = -4 + 12 = 8 \quad (\text{increasing})
f′(1)=4(1)3−12(1)=4−12=−8(decreasing) f'(1) = 4(1)^3 - 12(1) = 4 - 12 = -8 \quad (\text{decreasing})
f′(2)=4(2)3−12(2)=32−24=8(increasing) f'(2) = 4(2)^3 - 12(2) = 32 - 24 = 8 \quad (\text{increasing})
Intervals: - Decreasing on (−∞,−3) (-\infty, -\sqrt{3}) and (0,3) (0, \sqrt{3}) . - Increasing on (−3,0) (-\sqrt{3}, 0) and (3,∞) (\sqrt{3}, \infty) .

STEP 5

Find the local maximum and minimum values of f f :
- Local minima occur where f′ f' changes from decreasing to increasing. - Local maxima occur where f′ f' changes from increasing to decreasing.
From intervals: - Local minimum at x=−3 x = -\sqrt{3} and x=3 x = \sqrt{3} . - Local maximum at x=0 x = 0 .
Evaluate f(x) f(x) at these points:
f(−3)=(−3)4−6(−3)2=9−18=−9 f(-\sqrt{3}) = (-\sqrt{3})^4 - 6(-\sqrt{3})^2 = 9 - 18 = -9
f(3)=(3)4−6(3)2=9−18=−9 f(\sqrt{3}) = (\sqrt{3})^4 - 6(\sqrt{3})^2 = 9 - 18 = -9
f(0)=04−6(0)2=0 f(0) = 0^4 - 6(0)^2 = 0
Local minima: f(−3)=−9 f(-\sqrt{3}) = -9 , f(3)=−9 f(\sqrt{3}) = -9 .
Local maximum: f(0)=0 f(0) = 0 .

STEP 6

Find the inflection points by finding the second derivative and setting it equal to zero:
f′′(x)=ddx(4x3−12x)=12x2−12 f''(x) = \frac{d}{dx}(4x^3 - 12x) = 12x^2 - 12
Set the second derivative equal to zero:
12x2−12=0 12x^2 - 12 = 0
12(x2−1)=0 12(x^2 - 1) = 0
x2−1=0 x^2 - 1 = 0
x=±1 x = \pm 1
Inflection points occur at x=1,−1 x = 1, -1 .

STEP 7

Determine the intervals of concavity by testing intervals around the inflection points in the second derivative:
Test intervals: (−∞,−1),(−1,1),(1,∞) (-\infty, -1), (-1, 1), (1, \infty) .
Choose test points: x=−2,0,2 x = -2, 0, 2 .
Evaluate f′′(x) f''(x) at these points:
f′′(−2)=12(−2)2−12=48−12=36(concave up) f''(-2) = 12(-2)^2 - 12 = 48 - 12 = 36 \quad (\text{concave up})
f′′(0)=12(0)2−12=−12(concave down) f''(0) = 12(0)^2 - 12 = -12 \quad (\text{concave down})
f′′(2)=12(2)2−12=48−12=36(concave up) f''(2) = 12(2)^2 - 12 = 48 - 12 = 36 \quad (\text{concave up})
Intervals: - Concave up on (−∞,−1) (-\infty, -1) and (1,∞) (1, \infty) . - Concave down on (−1,1) (-1, 1) .
The critical numbers are x=0,3,−3 x = 0, \sqrt{3}, -\sqrt{3} . The function is increasing on (−3,0) (-\sqrt{3}, 0) and (3,∞) (\sqrt{3}, \infty) , and decreasing on (−∞,−3) (-\infty, -\sqrt{3}) and (0,3) (0, \sqrt{3}) . Local minima are f(−3)=−9 f(-\sqrt{3}) = -9 and f(3)=−9 f(\sqrt{3}) = -9 . The local maximum is f(0)=0 f(0) = 0 . Inflection points are at x=1,−1 x = 1, -1 . The function is concave up on (−∞,−1) (-\infty, -1) and (1,∞) (1, \infty) , and concave down on (−1,1) (-1, 1) .

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