Science:Math Exam Resources/Courses/MATH100/December 2011/Question 08
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Question 08 

FullSolution Problems. In questions 28, justify your answers and show all your work. If a box is provided, write your final answer there. Simplification of answers is not required unless explicitly stated. If y = ƒ(x) is a continuous function with domain [0,1] and range in [3,5], prove that the line y = 2x + 3 intersects the graph of y = ƒ(x) at least once. 
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? 
If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint. 
Hint 1 

What theorem have you learned that applies to functions that are continuous on a closed interval? How could you use it for this problem? 
Hint 2 

The Intermediate Value Theorem states that if ƒ is a continuous function on a closed interval [a,b ], and N is any number between ƒ(a) and ƒ(b), there exists a c for which ƒ(c)=N. 
Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

Solution 

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Please rate my easiness! It's quick and helps everyone guide their studies. We want to use the intermediate value theorem, as suggested by the hints. To do this we have to combine the two given functions ƒ and y = 2x + 3 to one. So define for x in the interval [0,1]. The function ƒ crosses the line y = 2x + 3 when both functions are equal, i.e. at zeros of the function h. Since we are asked to show they cross at least one, we have to show that the function h has at least one zero. In other words, we need to show that there exists a value c in [0,1] such that h(c) = 0. To show this we need to make use of the fact that the range of the function ƒ is in [3,5], i.e. for any value of x in the domain [0,1] of the function ƒ. This implies that and Now, in the case that either h(0) or h(1) actually takes the value 0, we are done because we have found a zero of h. In the case that neither takes the value zero, then we know that This is where the Intermediate Value Theorem shines: h is a continuous function on [0,1], hence for any value N between h(0) and h(1) there exists a number c = c(N) such that Since we have shown that h(0) is negative and h(1) is positive, we can choose that value N to be exactly zero and thus the Intermediate Value Theorem guarantees the existence of a number c in the interval [0,1] such that h(c) = 0. Conclusion: At this value c we have that or, in other words, f(c) = 2c+3, which is what we had to show. 