Question Number 26381 by Tinkutara last updated on 24/Dec/17
Commented by Tinkutara last updated on 24/Dec/17
Should answer be (4) or not?
Commented by mrW1 last updated on 24/Dec/17
$${I}\:{think}\:\mathrm{1}\:{and}\:\mathrm{4}\:{are}\:{true}. \\ $$
Commented by Tinkutara last updated on 24/Dec/17
But answer given is only (1).
Commented by prakash jain last updated on 24/Dec/17
$${y}'=\frac{{dy}}{{dx}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\underset{{h}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{{y}'\left({x}+{h}\right)−{y}'}{{h}}\:\:\left({I}\right) \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\underset{{h}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{{y}'\left({x}\right)−{y}'\left({x}−{h}\right)}{{h}}\:\left({II}\right) \\ $$$${if}\:{y}'\:{is}\:{increasing}\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }>\mathrm{0} \\ $$
Commented by mrW1 last updated on 25/Dec/17
$${let}'{s}\:{look}\:{at}\:{an}\:{example}: \\ $$$${y}'=\left({x}−\mathrm{3}\right)^{\mathrm{3}} +\mathrm{2} \\ $$$${it}\:{is}\:{increasing}\:{for}\:{all}\:{x}. \\ $$$${but}\:{y}''=\mathrm{3}\left({x}−\mathrm{3}\right)^{\mathrm{2}} \geqslant\mathrm{0}\:{not}\:>\mathrm{0} \\ $$$${that}\:{means}\:{if}\:{y}'\:{is}\:{increasing}, \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:{could}\:{also}\:{be}\:{equal}\:{to}\:{zero}. \\ $$
Commented by mrW1 last updated on 25/Dec/17
$${so}\:{the}\:{answer}\:{is}\:{indeed}\:{only}\:\left(\mathrm{1}\right), \\ $$$${since}\:{at}\:{x}=\mathrm{0},\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:{could}\:{be}\:=\mathrm{0}. \\ $$$${for}\:{example}\:{if}\:{the}\:{curve}\:{is} \\ $$$${y}={x}^{{n}} \:{with}\:{n}\geqslant\mathrm{3}. \\ $$
Commented by prakash jain last updated on 25/Dec/17
$$\mathrm{correct}.\:\mathrm{For}\:\mathrm{strictly}\:\mathrm{increasing} \\ $$$$\mathrm{function}\:\mathrm{derivate}\:\mathrm{is}\:\mathrm{greater}\:\mathrm{than} \\ $$$$\mathrm{or}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{0}. \\ $$
Commented by mrW1 last updated on 25/Dec/17
$${thanks}\:{for}\:{checking}\:{sir}!\:{you}\:{made} \\ $$$${and}\:{make}\:{a}\:{great}\:{contribution}\:{to}\:{this}\: \\ $$$${community}. \\ $$
Commented by Tinkutara last updated on 25/Dec/17
Thanks to both mrW1 and prakash jain!