Function and graph

Calculus Level pending

$y= x^3+3|x|+c$ is a function.

Which of the following statements is/are true?

1 . If $-2\leq c\leq 0$ , then $y$ has at least two roots.
2 . If $0\leq c\leq 2$ , then y has at most one root.
3 . If $f'(b)=0$ , $b$ is the root of $y$ , then $c = 2$ .
4 . $y$ is decreasing in $y \in (a,b)$ and $b-a=2$ .

Statement 3 and Statement 4 Statement 1, Statement 2 and Statement 4 Statement 2 and Statement 3 Statement 1 and Statement 4

1 solution

Akash Shukla
Apr 30, 2016

$y= x^3+3|x|+c$ . (1)

*Case :- 1 *

if x≥0,

$y= x^3+3x+c$

∴ dy/dx = $3x^2+3$ = 3( $x^2+1)$ > 0

hence for every x>0 , y is increasing.

'y' will have it's intercept on y-axis at (0,c).

Case :- 2

if x<0,

$y= x^3-3x+c$

dy/dx= $3x^2-3$ = 3( $x^2-1)$

y is decreasing for |x|<1

hence for -1<x<0, y is decreasing.

dy/dx = 0 ⇒ x=-1 and y=c+2

so at (-1,c+2) the slope is zero.

also in x∈ (-∝,-1), y is increasing.

so in y∈ (c,c+2), y is decreasing and c+2-c=2

so if 'y' touches X-axis then slope will be '0', and so x=-1 and y=0

putting in eqn. (1) , we get c=-2

so towards left it is increasing and towards right it is decreasing upto x=0 and then for x>0 it is increasing

Graph of y will be in third quadrant and then it have an intercept at (0,-2) , and then it increases and will again intersects on X-axis at some point in 4th quadrant and then increases in first quadrant. so it will have only two roots. But if c<-2, then it will have only one root , as it cannot touch X-axis. Also if c>0 then for x>0, y won't have any root and for x<0 y have only one root. And for -2<x<0 , y will have three roots as it intersects X-axis two times.

Function and graph

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