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
