Given a function f, an interval [a,b] and a value V, find a value c in the interval so that f(c)=V. Apply the Intermediate Value Theorem.

Question

 

Round-off your answers to 2 decimal places, stating from the least to the highest, if there would be more than one.  For positive answers, do not state the + sign anymore.  For answers less than 1, precede them with 0, example 0.11

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Answer 1

No values or expressions have been given for f(x), a, b or V, and the image does not display. Therefore an example will be given to demonstrate the method using IVT.

EXAMPLE

Let f(x)=(x²+1)/(x²-1), a=2, b=4 and V=1.2. Note that this function is not continuous generally, but is continuous in the interval [a,b] and decreases as x increases from a to b.

f(a)=f(2)=5/3=1.67 approx, f(b)=f(4)=17/15=1.13 approx.

c must be between 2 and 4, so let c=the average of 2 and 4=(2+4)/2=3.

f(3)=10/8=1.25, which is greater than V. Therefore c must be between 3 and 4, because V is between f(3)=1.25 and f(4)=1.13. The next value for is the average of 3 and 4=3.5

f(3.5)=13.25/11.25=53/45=1.18 approx, which is less than V. So 3<c<3.5, and the next estimate for c=3.25, when f(c)=1.21>V, so 3.25<c<3.5, the next estimate being 3.375, for which f(c)=1.192 approx.

Next c=(3.25+3.375)2=3.3125, f(3.3125)=1.2005>V. This result is close to V so c≈3.31 to 2 decimal places. More accuracy can be obtained by continuing the process to the third decimal place and then rounding. Another way is to linearly interpolate the narrowing interval rather than simply average the endpoints, but that is beyond the usual application of IVT.

It helps to sketch a graph so as to aid in deciding successive trial values for c.

In this example, c=√11=3.3166 is accurate to 4 decimal places.