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.

NOTE: 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

(a) f(x) = x^​2 on [0,3], V = 2
(b) f(x) = sin [0,
π/2], V = 1/2

(a)   c = _____

edited

(a) f(c)=2 when f(x)=x² in [0,3]. So we know by IVT that c is between 0 and 3, because f(0)=0 and f(3)=9, and 2 lies between f(0) and f(3), that is, between 0 and 9.

We can break the interval [0,3] down into two halves: [0,1.5] and [1.5,3].

f(1.5)=2.25 which is a bit more than 2, so c is in [0,1.5]. Now we can reduce the interval again. Let’s divide it into 3 this time: [0,0.5], [0.5,1], [1,1.5] and f(0.5)=0.25, f(1)=1. So we know c is in [1,1.5] because 2 is between 1 and 2.25. If we divide the interval again into slices of 0.1, we find that c must be between 1.4 and 1.5 because f(1.4)=1.96 which is close to 2 and 2.25 is close to 2. We may have to use a calculator to get even closer. We could split [1.4,1.5] into two: [1.40,1.45] and [1.45,1.50] as we move towards finding the second decimal place. The calculator gives us 1.45²=2.1025, so f(1.45)=2.1025, which is a bit more than 2 so c is in [1.40,1.45]. And we can break this down further into 0.01 slices. Let’s take the interval [1.41,1.42] and see what we get: f(1.41)=1.9881 and f(1.42)=2.0164. So we know c is in [1.41,1.42] because 2 is between f(1.41) and f(1.42), but which value is nearer? 2-f(1.41)=0.0119 and f(1.42)-2=0.0164. So 2 is nearer to f(1.41) therefore c=1.41 to 2 decimal places.

(b) We apply the same logic as in (a). f(0)=sin(0)=0, f(π/2)=1. Note that π radians is equivalent to 180° so π/2 radians is equivalent to 90°. π is just a number, 3.1415... that goes on for ever but we only need 2 decimal places, so for us in this question we can use 3.14 and forget about π. So we are looking for c in [0,1.57].

We can split the range up into slices. For the sake of simplicity let’s use the range [0,1.5] because f(1.5)=sin(1.5 radians)=0.9975 and 0.5 lies between 0 (sin(0)=0) and 0.9975 (sin(1.5)=0.9975).

f(0.5)=0.4794 and sin(1)=0.8415 so 0.5 lies between these numbers so we know c lies between 0.5 and 1.

It turns out that in the next decimal place f(0.52)=0.4969 and f(0.53)=0.5055, so c is between 0.52 and 0.53, and 0.5 is nearer to 0.4969 than it is to 0.5055, so c=0.52 to 2 decimal places.

This is how IVT is applied. I use it to find the roots of equations which cannot be solved in any other way. The calculator does all the work.

[The answer to (a) is of course √2 precisely and to (b) is π/6, which is the same as 180/6=30°. But you were asked to use IVT and that’s just what we’ve done.]

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