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

 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

Answers:

(a)  c = _____ ;  c = _____

(b)  c = _____

Answer 1

(a) f(0)=0 and f(3)=9, and 0<V<9 so 0<c<3.

Let's go for c=1.5 the centre of the interval: f(1.5)=2.25>V when V=2.

Let's now go for c=1.4 in an effort to "pinch" the root: f(1.4)=1.96<V, so 1.4<c<1.5.

We have c correct to 1 dec place. Let's go for the 2nd dec place by putting c=1.45; f(1.45)=2.1025>V and we already know that f(1.4)<V; we now know 1.4<c<1.45, so let's try an intermediate value c=1.42, f(1.42)>V. And now put c=1.41 to pinch the root: f(1.41)<V so we know 1.41<c<1.42. Both values are within [0,3] so we have the upper and lower limits for c. We don't have to consider negative values for c because the interval is positive in scope.

(b) f(x)=sin(x) on [0,π/2]. sin(0)=0 and sin(π/2)=1. V=1/2=0.5, which is halfway in the range so let's put c=π/4. f(π/4)=0.7071>V; try c=π/8, f(π/8)=0.3827<V. Now try c=π/6 which lies between π/4 and π/8, f(π/6)=0.5=V. So we have the exact root c=π/6.

 

Comments

Hi, Rod,

Thanks for your answer. But still I'm confused.

Could you please provide the value of c only as per required in the blank?

Thanks

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For (a) I would put c=1.41 and c=1.42 and for (b) I would put c=π/6. In (a) I think they want you to specify an interval for c; but in (b) it's an exact value.