Let f be a derivable function on R such that its derivative function is given by f prime(x)= (4 x2-x4-1/2(x4+1)), then

Question

Let f be a derivable function on R such that its derivative function is given by f prime(x)= (4 x2-x4-1/2(x4+1)), then (A) f is concave up for x ∈(-∞,-1) ∪(0,1) (B) f has exactly one point of local minimum. (C) f has an inflexion point at x =-1,0,1. (D) fhas exactly two points of local maximum.

Tardigrade Admin · Accepted Answer

f prime( x )=(2 x 2/(1+ x 4))-(1/2) Let g(x)=(2 x2/1+x4) then g prime(x)=(-4 x(x-1)(x+1)/(1+x4)2)

Q. Let $f$ be a derivable function on $R$ such that its derivative function is given by $f^{'} (x) =$ $\frac{4 x ^{2} - x ^{4} - 1}{2 ( x ^{4} + 1 )}$ , then

$f$ is concave up for $x \in (- \infty, - 1) \cup (0, 1)$

$f$ has exactly one point of local minimum.

$f$ has an inflexion point at $x = - 1, 0, 1$ .

$f$ has exactly two points of local maximum.

$f^{'} (x) = \frac{2 x ^{2}}{( 1 + x ^{4} )} - \frac{1}{2}$
Let $g (x) = \frac{2 x ^{2}}{1 + x ^{4}}$
then $g^{'} (x) = \frac{- 4 x ( x - 1 ) ( x + 1 )}{( 1 + x ^{4} ) ^{2}}$

Q. Let f be a derivable function on R such that its derivative function is given by f′(x)= 2(x4+1)4x2−x4−1​, then

f is concave up for x∈(−∞,−1)∪(0,1)

f has exactly one point of local minimum.

f has an inflexion point at x=−1,0,1.

fhas exactly two points of local maximum.