`f(x)=x^(1/5)` at `x=0` has inflection point. `y"\ DdotNdotEdot\ ` at `x=0`

f(x)=x^(3)atx=0 has inflection point y'=0 at x=0 and change sign.

f(x)=|x^(2)-1| has no inflection point in its domain as no tangent can be drawn at these points.

The function f(x)=x^((1)/(3))(x-1) has two inflection points has one point of extremum is non-differentiable has range [-3x2^(-(8)/(3)),oo)

The curve f(x)=x^(2/3) ,at the point with x=0 ,has

Concavity and convexity : if f''(x) gt 0 AA x in (a,b) then the curve y=f(x) is concave up ( or convex down) in (a,b) and if f''(x) lt 0 AA x in (a,b) then the curve y=f(x) is concave down (or convex up ) in (a,b) Inflection point : The point where concavity of the curve changes is known as point of inflection (at inflection point f''(x) is equal to 0 or undefined) Number of point of inflection for f(x) =(x-1)^(3) (x-2)^(2) is :

Concavity and convexity : if f''(x) gt 0 AA x in (a,b) then the curve y=f(x) is concave up ( or convex down) in (a,b) and if f''(x) lt 0 AA x in (a,b) then the curve y=f(x) is concave down (or convex up ) in (a,b) Inflection point : The point where concavity of the curve changes is known as point of inflection (at inflection point f''(x) is equal to 0 or undefined) Exhaustive set of values of 'a' for which the function f(x) =x^(4) +ax^(3)+(3x^(2))/(2)+1 will be concave upward along the entire real line is :

A function y-f(x) has a second-order derivative f'(x)=6(x-1). It its graph passes through the point (2,1) and at that point tangent to the graph is y=3x-5 , then the value of f(0) is 1 (b) -1(c)2 (d) 0

For a cubic function y=f(x),f'(x)=4x at each point (x,y) on it and it crosses the x- axis at (-2,0) at an angle of 45^(@) with positive direction of the x -axis.Then the value of (f(1))/(5)| is