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Certainly! Let's work through the given function [tex]\( f(x) = \frac{1}{3} x^3 - 2 x^2 + 3 x + 7 \)[/tex] step by step.
### i) Find [tex]\( f'(x) \)[/tex]
To find [tex]\( f'(x) \)[/tex], we need to differentiate [tex]\( f(x) \)[/tex] with respect to [tex]\( x \)[/tex].
[tex]\[ f'(x) = \frac{d}{dx}\left(\frac{1}{3} x^3 - 2 x^2 + 3 x + 7\right) \][/tex]
Differentiating each term individually:
[tex]\[ \frac{d}{dx}\left(\frac{1}{3} x^3\right) = x^2 \][/tex]
[tex]\[ \frac{d}{dx}\left(-2 x^2\right) = -4 x \][/tex]
[tex]\[ \frac{d}{dx}\left(3 x\right) = 3 \][/tex]
[tex]\[ \frac{d}{dx}\left(7\right) = 0 \][/tex]
So, we get:
[tex]\[ f'(x) = x^2 - 4x + 3 \][/tex]
### ii) Find the coordinates of the turning points of [tex]\( f(x) \)[/tex]
Turning points occur where [tex]\( f'(x) = 0 \)[/tex]. We need to solve for [tex]\( x \)[/tex] in the equation:
[tex]\[ x^2 - 4x + 3 = 0 \][/tex]
Factoring the quadratic:
[tex]\[ (x - 1)(x - 3) = 0 \][/tex]
Setting each factor to zero gives us the critical points:
[tex]\[ x = 1 \][/tex]
[tex]\[ x = 3 \][/tex]
To find the coordinates of these turning points, we substitute these [tex]\( x \)[/tex]-values back into the original function [tex]\( f(x) \)[/tex]:
For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \frac{1}{3}(1)^3 - 2(1)^2 + 3(1) + 7 = \frac{1}{3} - 2 + 3 + 7 = 8.3333 \][/tex]
For [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = \frac{1}{3}(3)^3 - 2(3)^2 + 3(3) + 7 = 9 - 18 + 9 + 7 = 7 \][/tex]
Thus, the turning points are:
[tex]\[ (1, 8.3333) \][/tex]
[tex]\[ (3, 7) \][/tex]
### iii) State the nature of the turning points in (ii) above
To determine the nature of these turning points, we look at the second derivative [tex]\( f''(x) \)[/tex].
First, find [tex]\( f''(x) \)[/tex]:
[tex]\[ f''(x) = \frac{d}{dx}(x^2 - 4x + 3) = 2x - 4 \][/tex]
Now, evaluate [tex]\( f''(x) \)[/tex] at the critical points:
For [tex]\( x = 1 \)[/tex]:
[tex]\[ f''(1) = 2(1) - 4 = -2 \][/tex]
Since [tex]\( f''(1) < 0 \)[/tex], the point [tex]\( (1, 8.3333) \)[/tex] is a local maximum.
For [tex]\( x = 3 \)[/tex]:
[tex]\[ f''(3) = 2(3) - 4 = 2 \][/tex]
Since [tex]\( f''(3) > 0 \)[/tex], the point [tex]\( (3, 7) \)[/tex] is a local minimum.
### iv) Find the values of [tex]\( x \)[/tex] for which [tex]\( f'(x) > 0 \)[/tex]
We need to solve the inequality:
[tex]\[ x^2 - 4x + 3 > 0 \][/tex]
This can be factored as:
[tex]\[ (x - 1)(x - 3) > 0 \][/tex]
The solution to this inequality:
[tex]\[ x < 1 \quad \text{or} \quad x > 3 \][/tex]
Expressing in interval notation:
[tex]\[ (-\infty, 1) \cup (3, \infty) \][/tex]
### v) Find the values of [tex]\( x \)[/tex] for which [tex]\( f'(x) < 0 \)[/tex]
We need to solve the inequality:
[tex]\[ x^2 - 4x + 3 < 0 \][/tex]
The solution to this inequality:
[tex]\[ 1 < x < 3 \][/tex]
Expressing in interval notation:
[tex]\[ (1, 3) \][/tex]
### Summary
- [tex]\( f'(x) = x^2 - 4x + 3 \)[/tex]
- Turning points are [tex]\( (1, 8.3333) \)[/tex] and [tex]\( (3, 7) \)[/tex]
- The point [tex]\( (1, 8.3333) \)[/tex] is a local maximum, and the point [tex]\( (3, 7) \)[/tex] is a local minimum
- [tex]\( f'(x) > 0 \)[/tex] for [tex]\( x \in (-\infty, 1) \cup (3, \infty) \)[/tex]
- [tex]\( f'(x) < 0 \)[/tex] for [tex]\( x \in (1, 3) \)[/tex]
### i) Find [tex]\( f'(x) \)[/tex]
To find [tex]\( f'(x) \)[/tex], we need to differentiate [tex]\( f(x) \)[/tex] with respect to [tex]\( x \)[/tex].
[tex]\[ f'(x) = \frac{d}{dx}\left(\frac{1}{3} x^3 - 2 x^2 + 3 x + 7\right) \][/tex]
Differentiating each term individually:
[tex]\[ \frac{d}{dx}\left(\frac{1}{3} x^3\right) = x^2 \][/tex]
[tex]\[ \frac{d}{dx}\left(-2 x^2\right) = -4 x \][/tex]
[tex]\[ \frac{d}{dx}\left(3 x\right) = 3 \][/tex]
[tex]\[ \frac{d}{dx}\left(7\right) = 0 \][/tex]
So, we get:
[tex]\[ f'(x) = x^2 - 4x + 3 \][/tex]
### ii) Find the coordinates of the turning points of [tex]\( f(x) \)[/tex]
Turning points occur where [tex]\( f'(x) = 0 \)[/tex]. We need to solve for [tex]\( x \)[/tex] in the equation:
[tex]\[ x^2 - 4x + 3 = 0 \][/tex]
Factoring the quadratic:
[tex]\[ (x - 1)(x - 3) = 0 \][/tex]
Setting each factor to zero gives us the critical points:
[tex]\[ x = 1 \][/tex]
[tex]\[ x = 3 \][/tex]
To find the coordinates of these turning points, we substitute these [tex]\( x \)[/tex]-values back into the original function [tex]\( f(x) \)[/tex]:
For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \frac{1}{3}(1)^3 - 2(1)^2 + 3(1) + 7 = \frac{1}{3} - 2 + 3 + 7 = 8.3333 \][/tex]
For [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = \frac{1}{3}(3)^3 - 2(3)^2 + 3(3) + 7 = 9 - 18 + 9 + 7 = 7 \][/tex]
Thus, the turning points are:
[tex]\[ (1, 8.3333) \][/tex]
[tex]\[ (3, 7) \][/tex]
### iii) State the nature of the turning points in (ii) above
To determine the nature of these turning points, we look at the second derivative [tex]\( f''(x) \)[/tex].
First, find [tex]\( f''(x) \)[/tex]:
[tex]\[ f''(x) = \frac{d}{dx}(x^2 - 4x + 3) = 2x - 4 \][/tex]
Now, evaluate [tex]\( f''(x) \)[/tex] at the critical points:
For [tex]\( x = 1 \)[/tex]:
[tex]\[ f''(1) = 2(1) - 4 = -2 \][/tex]
Since [tex]\( f''(1) < 0 \)[/tex], the point [tex]\( (1, 8.3333) \)[/tex] is a local maximum.
For [tex]\( x = 3 \)[/tex]:
[tex]\[ f''(3) = 2(3) - 4 = 2 \][/tex]
Since [tex]\( f''(3) > 0 \)[/tex], the point [tex]\( (3, 7) \)[/tex] is a local minimum.
### iv) Find the values of [tex]\( x \)[/tex] for which [tex]\( f'(x) > 0 \)[/tex]
We need to solve the inequality:
[tex]\[ x^2 - 4x + 3 > 0 \][/tex]
This can be factored as:
[tex]\[ (x - 1)(x - 3) > 0 \][/tex]
The solution to this inequality:
[tex]\[ x < 1 \quad \text{or} \quad x > 3 \][/tex]
Expressing in interval notation:
[tex]\[ (-\infty, 1) \cup (3, \infty) \][/tex]
### v) Find the values of [tex]\( x \)[/tex] for which [tex]\( f'(x) < 0 \)[/tex]
We need to solve the inequality:
[tex]\[ x^2 - 4x + 3 < 0 \][/tex]
The solution to this inequality:
[tex]\[ 1 < x < 3 \][/tex]
Expressing in interval notation:
[tex]\[ (1, 3) \][/tex]
### Summary
- [tex]\( f'(x) = x^2 - 4x + 3 \)[/tex]
- Turning points are [tex]\( (1, 8.3333) \)[/tex] and [tex]\( (3, 7) \)[/tex]
- The point [tex]\( (1, 8.3333) \)[/tex] is a local maximum, and the point [tex]\( (3, 7) \)[/tex] is a local minimum
- [tex]\( f'(x) > 0 \)[/tex] for [tex]\( x \in (-\infty, 1) \cup (3, \infty) \)[/tex]
- [tex]\( f'(x) < 0 \)[/tex] for [tex]\( x \in (1, 3) \)[/tex]
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