Urgent help needed, calculus applications (1 Viewer)

[Shoom]

Find the staionary points on the curve y= (3x-1)(x-2)^4

Differentiate y=xsqrtx+1

the curve fx = ax^4-2x^3+7x^2-x+5 has a staionary point at x=1 find a


show that sqrt of x has no staionary points


Thanks

please show working

 

[pLuvia]

1.
y'=3(3x-1)(x-2)^4+4(3x-1)(x-2)^3=0
(3x-1)(x-2)^3[3(x-2)+4]=0

then solve for x and plug those x's back into y then you get your stat points

2. xsqrtx=x^(3/2)
y'=(3/2)x^1/2

3. f'=4ax^3-6x^2+14x-1
x=1 is a stat point so when x=1 f'=0
f'(1)=4a-6-14-1=0
a=21/4

4. y=x^0.5
y'=0.5x^-0.5
Let y'=0
0.5x^-0.5=0

The only way for y' to be equal to zero is if x is zero but if x is zero the function is undefined hence there is no stationary points