Answer :
Sure, let's analyze the function [tex]\( f(x) = \frac{1}{3} x^4 - 5 x^2 + 12 \)[/tex] step-by-step.
### Critical Points
1. Find the first derivative [tex]\( f'(x)\)[/tex] to locate the critical points:
[tex]\[ f'(x) = \frac{d}{dx} \left( \frac{1}{3} x^4 - 5 x^2 + 12 \right) \][/tex]
[tex]\[ f'(x) = \frac{4}{3} x^3 - 10x \][/tex]
2. Set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ \frac{4}{3} x^3 - 10x = 0 \][/tex]
Factor out the common term:
[tex]\[ x \left( \frac{4}{3} x^2 - 10 \right) = 0 \][/tex]
This gives us two equations to solve:
[tex]\[ x = 0 \][/tex]
[tex]\[ \frac{4}{3} x^2 = 10 \][/tex]
Solving the second equation:
[tex]\[ x^2 = \frac{30}{4} = 7.5 \][/tex]
Taking the square root of both sides:
[tex]\[ x = \pm \sqrt{7.5} \][/tex]
Approximating the square root:
[tex]\[ x \approx \pm 2.7386 \][/tex]
So, the critical points are:
[tex]\[ x = -2.7386, \; 0, \; 2.7386 \][/tex]
### Concavity at Critical Points
1. Find the second derivative [tex]\( f''(x)\)[/tex]:
[tex]\[ f''(x) = \frac{d}{dx} \left( \frac{4}{3} x^3 - 10x \right) \][/tex]
[tex]\[ f''(x) = 4x^2 - 10 \][/tex]
2. Evaluate the second derivative at each critical point to determine concavity:
- At [tex]\( x = -2.7386 \)[/tex]:
[tex]\[ f''(-2.7386) = 4(-2.7386)^2 - 10 \approx 20 \][/tex]
- At [tex]\( x = 0 \)[/tex]:
[tex]\[ f''(0) = 4(0)^2 - 10 = -10 \][/tex]
- At [tex]\( x = 2.7386 \)[/tex]:
[tex]\[ f''(2.7386) = 4(2.7386)^2 - 10 \approx 20 \][/tex]
Interpretation of concavity:
- [tex]\( f(x) \)[/tex] is concave up (minimum) at [tex]\( x = -2.7386 \)[/tex] and [tex]\( x = 2.7386 \)[/tex] because [tex]\( f''(x) > 0 \)[/tex].
- [tex]\( f(x) \)[/tex] is concave down (maximum) at [tex]\( x = 0 \)[/tex] because [tex]\( f''(x) < 0 \)[/tex].
### Points of Inflection
1. Set the second derivative equal to zero to find points of inflection:
[tex]\[ f''(x) = 4x^2 - 10 = 0 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ 4x^2 = 10 \][/tex]
[tex]\[ x^2 = 2.5 \][/tex]
Taking the square root of both sides:
[tex]\[ x = \pm \sqrt{2.5} \][/tex]
Approximate the square root:
[tex]\[ x \approx \pm 1.5811 \][/tex]
2. Determine the corresponding [tex]\( y \)[/tex]-values for the inflection points by substituting [tex]\( x \)[/tex] back into [tex]\( f(x) \)[/tex]:
- For [tex]\( x = 1.5811 \)[/tex]:
[tex]\[ f(1.5811) \approx \frac{1}{3} (1.5811)^4 - 5 (1.5811)^2 + 12 \approx 1.5833 \][/tex]
- For [tex]\( x = -1.5811 \)[/tex]:
[tex]\[ f(-1.5811) \approx \frac{1}{3} (-1.5811)^4 - 5 (-1.5811)^2 + 12 \approx 1.5833 \][/tex]
The points of inflection are:
[tex]\[ (-1.5811, 1.5833) \quad \text{and} \quad (1.5811, 1.5833) \][/tex]
### Summary
- Critical points: [tex]\(-2.7386, 0, 2.7386\)[/tex]
- Concavity at critical points:
- [tex]\( x = -2.7386 \)[/tex]: Concave up (Minimum)
- [tex]\( x = 0 \)[/tex]: Concave down (Maximum)
- [tex]\( x = 2.7386 \)[/tex]: Concave up (Minimum)
- Inflection points and their values:
- [tex]\((-1.5811, 1.5833)\)[/tex]
- [tex]\((1.5811, 1.5833)\)[/tex]
### Critical Points
1. Find the first derivative [tex]\( f'(x)\)[/tex] to locate the critical points:
[tex]\[ f'(x) = \frac{d}{dx} \left( \frac{1}{3} x^4 - 5 x^2 + 12 \right) \][/tex]
[tex]\[ f'(x) = \frac{4}{3} x^3 - 10x \][/tex]
2. Set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ \frac{4}{3} x^3 - 10x = 0 \][/tex]
Factor out the common term:
[tex]\[ x \left( \frac{4}{3} x^2 - 10 \right) = 0 \][/tex]
This gives us two equations to solve:
[tex]\[ x = 0 \][/tex]
[tex]\[ \frac{4}{3} x^2 = 10 \][/tex]
Solving the second equation:
[tex]\[ x^2 = \frac{30}{4} = 7.5 \][/tex]
Taking the square root of both sides:
[tex]\[ x = \pm \sqrt{7.5} \][/tex]
Approximating the square root:
[tex]\[ x \approx \pm 2.7386 \][/tex]
So, the critical points are:
[tex]\[ x = -2.7386, \; 0, \; 2.7386 \][/tex]
### Concavity at Critical Points
1. Find the second derivative [tex]\( f''(x)\)[/tex]:
[tex]\[ f''(x) = \frac{d}{dx} \left( \frac{4}{3} x^3 - 10x \right) \][/tex]
[tex]\[ f''(x) = 4x^2 - 10 \][/tex]
2. Evaluate the second derivative at each critical point to determine concavity:
- At [tex]\( x = -2.7386 \)[/tex]:
[tex]\[ f''(-2.7386) = 4(-2.7386)^2 - 10 \approx 20 \][/tex]
- At [tex]\( x = 0 \)[/tex]:
[tex]\[ f''(0) = 4(0)^2 - 10 = -10 \][/tex]
- At [tex]\( x = 2.7386 \)[/tex]:
[tex]\[ f''(2.7386) = 4(2.7386)^2 - 10 \approx 20 \][/tex]
Interpretation of concavity:
- [tex]\( f(x) \)[/tex] is concave up (minimum) at [tex]\( x = -2.7386 \)[/tex] and [tex]\( x = 2.7386 \)[/tex] because [tex]\( f''(x) > 0 \)[/tex].
- [tex]\( f(x) \)[/tex] is concave down (maximum) at [tex]\( x = 0 \)[/tex] because [tex]\( f''(x) < 0 \)[/tex].
### Points of Inflection
1. Set the second derivative equal to zero to find points of inflection:
[tex]\[ f''(x) = 4x^2 - 10 = 0 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ 4x^2 = 10 \][/tex]
[tex]\[ x^2 = 2.5 \][/tex]
Taking the square root of both sides:
[tex]\[ x = \pm \sqrt{2.5} \][/tex]
Approximate the square root:
[tex]\[ x \approx \pm 1.5811 \][/tex]
2. Determine the corresponding [tex]\( y \)[/tex]-values for the inflection points by substituting [tex]\( x \)[/tex] back into [tex]\( f(x) \)[/tex]:
- For [tex]\( x = 1.5811 \)[/tex]:
[tex]\[ f(1.5811) \approx \frac{1}{3} (1.5811)^4 - 5 (1.5811)^2 + 12 \approx 1.5833 \][/tex]
- For [tex]\( x = -1.5811 \)[/tex]:
[tex]\[ f(-1.5811) \approx \frac{1}{3} (-1.5811)^4 - 5 (-1.5811)^2 + 12 \approx 1.5833 \][/tex]
The points of inflection are:
[tex]\[ (-1.5811, 1.5833) \quad \text{and} \quad (1.5811, 1.5833) \][/tex]
### Summary
- Critical points: [tex]\(-2.7386, 0, 2.7386\)[/tex]
- Concavity at critical points:
- [tex]\( x = -2.7386 \)[/tex]: Concave up (Minimum)
- [tex]\( x = 0 \)[/tex]: Concave down (Maximum)
- [tex]\( x = 2.7386 \)[/tex]: Concave up (Minimum)
- Inflection points and their values:
- [tex]\((-1.5811, 1.5833)\)[/tex]
- [tex]\((1.5811, 1.5833)\)[/tex]
