From the above definition we see the domain of the function is the set of all real numbers such that x≥−3 . Hence the domain is {x|x≥−3} .
(b)
To determine
To locate: any intercepts of the given function.
(b)
Expert Solution
Answer to Problem 34AYU
The function has a x intercept at (−52,0) .
And the y intercept is (0,−3) .
Explanation of Solution
Given information:
Given function
f(x)={2x+5 if−3≤x<0−3 ifx=0−5x ifx>0
Calculation:
The intercepts are the points on the graph which are obtained when it cuts the x and y axes. The points (−52,0)and(0,−3) satisfy the function above. Hence the function has a x intercept at (−52,0) .
And the y intercept is (0,−3) .
(c)
To determine
To sketch: the graph of the given function.
(c)
Expert Solution
Explanation of Solution
Given information:
Given function
f(x)={2x+5 if−3≤x<0−3 ifx=0−5x ifx>0
Calculation:
Plot the points and draw the line to get the graph of the function
(d)
To determine
To find: the range based on the graph.
(d)
Expert Solution
Answer to Problem 34AYU
The function f(x) is the set {y|y<5} .
Explanation of Solution
Given information:
Given function
f(x)={2x+5 if−3≤x<0−3 ifx=0−5x ifx>0
Calculation:
From the graph we see that f(x)<5 in its domain. So the range of the function f(x) is the set {y|y<5} .
(e)
To determine
To find: whether f continuous on its domain or not.
(e)
Expert Solution
Answer to Problem 34AYU
The graph it can be clearly seen that the function is discontinuous at the point x=0 .
Explanation of Solution
Given information:
Given function
f(x)={2x+5 if−3≤x<0−3 ifx=0−5x ifx>0
Calculation:
The only point at which the function might have behaved in a manner that it becomes discontinuous is x=0 . But at this point, the value of the function from the left of 0 and the right of 0 are given as:
f(0−)=2(0)+5=5
And
f(0)=−3
And
f(0+)=−5(0)=0
Hence
f(0+)≠f(0)≠f(0−)
So even at the break point the function is discontinuous.
Hence the function is discontinuous in its domain only at the point x=0 .
Also from the graph it can be clearly seen that the function is discontinuous at the point x=0 .
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.