Question

For y = -x2 + 2x there exist a c in the interval [- 19765, 19767] Such that f(c) = 0
(a) 1
(bl 2
(c) 3
(d) none of these​

Answer 1

Option c is currently correct answer in this question thank you,

Answer 2

Question: For $y=-x^2+2x$, there exist a $c$ in the interval $[-19765,19767]$ such that $f'(c)=0$ then find the value of $c$.

(a) 1          (b) 2        (c) 3        (d) None of these

Answer:

Option (a) is correct.

Step-by-step explanation:

Rolle's Theorem:-

If a function $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ such that $f(a)=f(b)$, then $f'(c) = 0$ for some $c$ with $a\leq c\leq b$.

Step 1 of 1

Consider the given function as follows:

$y=-x^2+2x$ . . . . (1)

Notice that the given function is a polynomial.

Thus, the function (1) is continuous in the interval $[-19765,19767]$.

Now,

Differentiate the function (1) as follows:

$\frac{dy}{dx} =\frac{d}{dx} (-x^2+2x)$

   $=-2x+2$

   $=2(1-x)$ . . . . . (2)

Clearly, the function (1) is differentiable in the interval $(-19765,19767)$.

Using the Rolle's theorem,

There exist a number $c$ such that $y'(c)=0$.

From (2), we have

$f'(x)=2(1-x)$

Replace $x$ by $c$ as follows:

$f'(c)=2(1-c)$

$0=2(1-c)$

⇒ $2(1-c)=0$ . . . . (3)

(a) 1

Substitute the value 1 for $c$ in the equation (3), we get

⇒ $2(1-1)=0$

⇒ $0=0$

Thus, option (a) is correct.

(b) 2

Substitute the value 2 for $c$ in the equation (3), we get

⇒ $2(1-2)=0$

⇒ $-2\neq 0$

Thus, option (b) is incorrect.

(c) 3

Substitute the value 3 for $c$ in the equation (3), we get

⇒ $2(1-3)=0$

⇒ $2(-2)=0$

⇒ $-4=0$

Thus, option (c) is incorrect.

Final answer: The value of $c$ is 1.

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