9. A function is given as [tex]f(x)=\frac{1}{3} x^3-2 x^2+3 x+7[/tex]. Find:

i) [tex]f^{\prime}(x)[/tex]

ii) The coordinates of the turning points of [tex]f(x)[/tex].

iii) State the nature of the turning points in (ii) above.

iv) The value(s) of [tex]x[/tex] for which [tex]f^{\prime}(x)\ \textgreater \ 0[/tex].

v) The value(s) of [tex]x[/tex] for which [tex]f^{\prime}(x)\ \textless \ 0[/tex].



Answer :

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]