in Calculus retagged by
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Let $f(x)$ be a real-valued function all of whose derivatives exist. Recall that a point $x_0$ in the domain is called an inflection point of $f(x)$ if the second derivative $f^{\prime \prime}(x)$ changes sign at $x_0$. Given the function $f(x)=\dfrac{x^5}{20}-\dfrac{x^4}{2}+3 x+1$, which of the following statement is true?

  1. $x_0=0$ is not an inflection point.
  2. $x_0=6$ is the only inflection point.
  3. $x_0=0$ and $x_0=6$, both are inflection points.
  4. The function does not have an inflection point.
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$ \large{\colorbox{yellow}{Detailed video solution of this question with direct time stamp}}$
All India Mock Test 4 - Solutions Part 1

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2 Answers

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4 votes
B is correct.

The point of this question is to check if the concept of the inflection point is understood; the given function will have one inflection point, but also points where the second derivative is zero but is not an inflection point.

Here $f^{\prime}(x)=\frac{x^4}{4}-2 x^3+3$ and $f^{\prime \prime}(x)=x^2(x-6)$, so $f^{\prime \prime}(x)$ vanishes at $0$ and $6;$ but only $6$ is an inflection point since $f^{\prime \prime \prime}(0)=0$ while $f^{\prime \prime \prime}(6) \neq 0$.
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4 Comments

to be point of inflection both f’(x) and f’’(x) should be equal to zero.
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inflection pt may or may not be critical pt
like y=x^3, x=0 is inflection pt, but at x=0, f`(x)$!=0$
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Thanks @DEBANJAN DAS2k !!

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4 votes
4 votes

Option A and B both should be the answer..

1 comment

Yes! correct!
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