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+5 votes

Consider the problem of maximizing $x^{2}-2x+5$ such that $0< x< 2$. The value of $x$ at which the maximum is achieved is:

  1. $0.5$
  2. $1$
  3. $1.5$
  4. $1.75$
  5. None of the above.
asked in Calculus by Boss (29.3k points)
edited by | 395 views

2 Answers

+12 votes
Best answer

$$P(x) = x^2 - 2x +5$$

Since a polynomial is defined and continuous everywhere, we only need to check the critical point and the boundaries.

$\dfrac{d}{dx}P(x) = 2x - 2$

Critical point: $2x-2 = 0\implies x = 1$ gives $P(x) = 4$, which is the minimum.

Boundaries: $\lim_{x \to \color{red}{0}^+} P(x) = \lim_{x \to \color{red}{2}^-} P(x) = 5$

Since $P(x)$ increases as $x$ goes farther away from the $1$, but $P(x)$ is defined on an open interval, $P(x)$ never attains a maximum!

Hence, e. None of the above is the correct answer.

answered by Boss (22.3k points)
Even there is no change in sign to left/rght of crtcl point no minima too.

plz revert @Akash
not getting plz explain in more simple way ?
For continuous functions, we need to see only the critical points (where the curve changes) to get the maxima and minima. Also, we should see the bundary points. But here, we have an open interval for the boundary and $f(x)$ increases as $x$ increases towards 2 and also as $x$ decreases towards 0. But $x$ can never be 0 or 2 as interval is open and not closed. This means we cannot say it is having a maxima at any point. Suppose $P(x) \geq 5, x = 1$, then 1 would have been the answer.
Sir can you explain what is "open interval for the boundary"? Is (0,2) not the boundaries?
0 votes

We could have check boundary conditions but given is an open interval.

answered by Active (4.5k points)

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