Question

Match the column

Answer 1

$f\left(x\right)=(x-1{)}^3(x+2{)}^5$
$\Rightarrow f\left(x\right)=3(x-1{)}^2(x+2{)}^5+5(x-1{)}^3(x+2{)}^4$
$\Rightarrow f^{\prime }\left(x\right)=(x-1{)}^2(x+2{)}^4\left[3\right(x+2)+5(x-1\left)\right]$
$=(x-1{)}^2(x+2{)}^2[8x+1]$
Sign of derivative does not change at $x=1$ and $x=-2$.
Sign of derivative changes sign at $x=-1/8$ from $-ve$ to $+ve$
Hence function has point of minima.
Also $f"\left(x\right)$ function has two points of inflection.
b. $f\left(x\right)=3\sin x+4\cos x-5x$
$\Rightarrow f^{\prime }\left(x\right)=3\cos x-4\sin x-5\le 0$, hence $f\left(x\right)$ is decreasing function
Also $f"\left(x\right)=-3\sin x-4\cos x=0$ for infinite values of $x$, hence function has infinite points of inflection.
c. From the graph $x=1$ point of maxima as well as point of inflection.
d. $f\left(x\right)(x-1{)}^{3/5}\Rightarrow f;(x)=\frac{3}{5}(x-1{)}^{2/5}\ge 0$ for all real $x$
Also, $f^{\prime }\left(x\right)=-\frac{3}{5}\frac{2}{5}(x-1{)}^{-7/5}$ which changes sign at $x=1$
Hence, $x=1$ is point of inflection.