5. Continuity on an Interval

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Presentation transcript:

5. Continuity on an Interval

Continuity definition revisited Remember the definition of continuity When asked about continuity, check all 3 things, then make a concluding statement

Example 1 Discuss the continuity of

Example 2 Find the value of a so that the function is continuous at x = -1.

Example 3 Find the value of a and b so that the function is continuous at x = 1.

Continuity on an interval A function f(x) is continuous on an open interval (a,b) if and only if the function f(x) is continuous at every point in the interval (a,b) A function f(x) is continuous on a closed interval [a,b] if and only if it is continuous on open interval (a,b) and if the endpoints exhibit the following: In words, this means there must be a closed dot.

Example 4 Revisiting the graph of f(x), graphus interruptus, determine whether the function is continuous on given intervals. If not, explain. A) (-5, -2) B) (-5,-2] C) [-5,-2) D) [-5,-2] E) [0,4] F) [6,8) G) (2,4) H) Find the largest value of b such that the function is continuous on (5,b] but not on (5,b+1]. i) Give the largest interval for a such that the function is continuous on (a,10).

Functions that are continuous for all values in their domain Polynomials – all real numbers Rational functions (VAs and holes not in domain) Root functions (where value under square root > 0) Trig functions (sin and cos – all real numbers, others not at VA) Exponential functions – all real numbers Log functions (where value of log > 0)

Intermediate value theorem A function that is continuous on a closed interval takes on every y value between f(a) and f(b).

In other words….. A function that is continuous can’t skip y values Note: Even though the function can’t skip y values, it can take on values that are outside of f(a) and f(b)

Example 5 Prove that sin x = 0.3 has at least one solution. We can apply the IVT because sin x is continuous Choose an interval (0, 𝜋 2 ) Sin(0) = 0, sin( 𝜋 2 ) = 1 0.3 is between these 2 values Therefore, the IVT tells us that sin x = 0.3 has a solution in (0, 𝜋 2 ) Could choose other intervals as well

Existence of zeros The IVT can be used to show the existence of zeros If one value of the function is negative and another is positive, there must be a zero somewhere in between

Example 6 Show that 𝑐𝑜𝑠 2 𝑥 −2𝑠𝑖𝑛 𝑥 4 has a zero in (0,2) f(0) = 1 f(2) = -0.786 Opposite signs, therefore a zero

Example 7 Here’s part of an example from the Free Response section of the 2007 AP exam. The functions f and g are differentiable (and continuous) for all real numbers, and g is strictly increasing. The function h is given by h(r)=f(g(r))-6. Explain why there must be a value r for 1<r<3 such that h(r) = -5.