in Calculus edited by
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6 votes
6 votes

Suppose $g(x)$ is a polynomial function such that $g(-1)=4$ and $g(2)=7$. Then there is a number $c$ between $-1$ and $2$ such that

  1. $g(c)=1$
  2. $g^{\prime}(c)=1$
  3. $g(c)=0$
  4. $g^{\prime}(c)=0$
in Calculus edited by
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3 Comments

Reason why Option B is correct.

We know 1 thing when g(x) is continuous & differentiable in [a, b] & (a, b).

Then g'(c) = g(b) - g(c) / (b-a)

In this question : g'(c) = 7 - 4/(2 - (-1)) = 3/3 = 1.

 

Option B is correct.
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5

No doubt for option B.

But for option-C:

As per g(x) is polynomial function and given g(-1)=4 and g(2)=7(we can say that these 2 points are connected) then there must be at least one point present such that g(c)=0.

Isn’t it?

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What if function is always increasing?
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1 Answer

5 votes
5 votes
Since $g(x)$ is a polynomial function it is continuous and differentiable everywhere. Therefore there are two theorems that we might use to answer this question. The first is the Intermediate Value Theorem, which says that between $-1$ and $2$ and any $y$-value between $4$ and $7$ there is at least one number $c$ such that $g(c)$ is equal to that $y$-value. Since none of the answer choices involve $y$-values between $4$ and $7$ , we go on to the next theorem.

The Mean Value Theorem says that between $-1$ and $2$ there is at least one number $c$ such that $g^{\prime}(c)$ is equal to the slope of the secant line between the points $(-1,4)$ and $(2,7)$, i.e. $1.$ So (B) is correct.
There is nothing else we can conclude from the information we are given.

2 Comments

edited by
Based on this reasoning, shouldn’t (C) be correct?
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@Arkaprava y value must have been between 4 and 7 so that Intermediate value theorem satisfies which is given as 0 hence No!

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