in Calculus retagged by
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3 votes
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Which of the following is/are TRUE?

  1. There is a differentiable function $f(x)$ with the property that $f(1)=-2$ and $f(5)=14$ and $f^{\prime}(x)\lt 3$ for every real number $x$.
  2. There exists a function $f$ such that $f(1)=-2, f(3)=0$, and $f^{\prime}(x)\gt 1$ for all $x$.
  3. For all functions $f$, if $a\lt b, f(a)\lt 0, f(b)\gt 0$, then there must be a number $c$, with $a\lt c\lt b$ and $f(c)=0$.
  4. If $f$ is differentiable at the number $x$, then it is continuous at $x$.
in Calculus retagged by
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f’(c ) = (f(b)-f(a))/(b-a)  as per Mean Value theorem.

(a) f’c = 4 so, false.

(b) f’c = 1 so, false.

© This option would have been correct if the given function f was given to be continuous in the interval [a,b]. But as it is not given, this is false.

(d) This is always a true statement as for function to be differentiable at any point, it should first  be continuous at that point.
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2 Answers

4 votes
4 votes

Check this $4$-minute video to get a detailed video solution -

https://www.youtube.com/watch?v=lWy-owo54wM&list=PLIPZ2_p3RNHjBd9WApLSvwrXnmZi2E7MV&index=34

edited by

1 comment

The above is indirect link.

Pls find direct link to the solution below

https://youtu.be/DPNNlYlPdlY?si=s773BQ4YzfZxVJsL
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2 votes
2 votes
1) It's given that $f(x)$ is differentiabile function, so it implies $f(x)$

is continuous function, the hypothesis for mean value was

satisfied, according to Lagrange's mvt if function $f(x)$ is

continuous on interval $[a,b]$ and differentiable on $(a,b)$ then

there exists point c in $(a,b)$ such that $f'(c) = \frac{f(b) -f(a)}{b-a}$

Now apply LMVT on interval [1,5] it gives us there exists point c in

$(1,5)$ such that $f'(c) = 4$, which makes the statement false.

 

2) Similar approach as above, it gives us there is c in (1,3) with

f'(c)= 1, which makes the statement 2 false.

 

3)Not necessarily, function can be discontinuous.

 

4)which is very obvious, so it's true.

2 Comments

Nice one!!!
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