#### To determine

a)

**To find:**

Values of x where local maximum and minimum occurs

#### Answer

Local minimum: at x=4, 8

Local maximum: at x=2, 6

#### Explanation

**1) Concept:**

i) If f’ changes from positive to negative at c, then f has a local maximum at c.

ii) If f’ changes from negative to positive at c, then f has a local minimum at c.

iii) If f’ is positive to the left and right of c or negative to the left and right of c, then f has no local maximum and minimum at c

**2) Calculation:**

By fundamental theorem,

g'x=f(x)

Therefore, the given graph is that of a derivative function

The local maxima and minima can be found where f’(x) =0, that is, where the graph meets the x axis

From the graph, the following conclusions can be made

On interval (0, 2) graph is increasing

On interval (2, 4) graph is decreasing

Since the graph is changing from increasing to decreasing at point x=2, it is a local maxima

On interval (2, 4) graph is decreasing

On interval (4, 6) graph is increasing

Since the graph is changing from decreasing to increasing at point 4, it is a local minima

From interval (4, 6) graph is increasing

From interval (6, 8) graph is decreasing

Since the graph is changing from increasing to decreasing at point 6, then it is local maxima

From interval (6, 8) graph is decreasing

From interval (8, 10) graph is increasing

Since the graph is changing from decreasing to increasing at point 8, then it is local minima

**Conclusion:**

Therefore,

Local minimum: at x=4, 8

Local maximum: at x=2, 6

#### To determine

b)

**To find:**

The absolute maximum value

#### Answer

x=2

#### Explanation

**1) Concept:**

Let c be a number in the domain D of a function f. Then f(c) is the

absolute maximum value of f on D if fc≥f(x) for all x in D

absolute minimum value of f on D if fc≤f(x) for all x in D

**2) Calculation:**

From interval (0,2) area of the shaded region can be calculated as

A=g(2)=∫02fdt

From interval (0,6) area of the shaded region can be calculated as

A=g6=∫06fdt=∫02fdt-∫24fdt+∫46fdt=g(2)-∫24fdt+∫46fdt

From interval (0,10) area of the shaded region can be calculated as

A=g(10)=∫010fdt=∫02fdt-∫24fdt+∫46fdt-∫68fdt+∫810fdt

=g6-∫68fdt+∫810fdt

Since the area of each region is decreasing it follows thatTherefore, g2>g6>g(10)

Hence, maximum value occurs at x=2

**Conclusion:**

Therefore, maximum value occurs at x=2

#### To determine

c)

**To find:**

On what interval is the curve concave downward

#### Answer

1,3, 5,7 and (9,10)

#### Explanation

**1) Concept:**

fx is concave upward if f''x>0 and fx is concave downward if f''x<0

f''x>0 means f’(x) is increasing and f’’(x)<0 means f’(x) is decreasing

**2) Calculation:**

Now, g’(x)=f(x), so g’’(x)=f’(x)

The graph of g will be concave downward where the above graph is decreasing.

From the graph g’(x) is decreasing on the interval, (1, 3), (5, 7) and (9, 10)

Therefore, g is concave downward on intervals, (1, 3),(5, 7) and (9, 10)

**Final Stetment:**

Therefore, interval is 1,3, 5,7 and (9,10)

#### To determine

d)

**To draw:**

Graph of g

#### Answer

#### Explanation

**Calculation:**

From the calculation performed in part ‘a’, there is local maximum at points 2, 6 and local minimum on points 4, 8. So draw a concave downward curve at point 2 and 6 and concave upward curve at point 4 and 8. Also maximum at point 2.

**Conclusion:**

The graph is