Is x= -4/3 a critical point of h(x) = (3x + 4)^5/4 ?

Answer 1

Holy crap! Misinformation!

h'(x)=(5/4)*(3x+4)^1/4 * (3)
h'(x)=(15/4)*(3x+4)^1/4
(15/4)*(3x+4)^1/4=0
(3x+4)^1/4=0
x=-4/3 is a critical value of h(x) since it is an x value that makes h'(x)=0

Answer 2

LOL. what conflicting answers! I hope this clears things up a little -

First, critical point probably refers to a point on the curve where the gradient of the tangent is 0. Thus as mentioned above by schleppin, x= -4/3 is not a critical point as the h'(-4/3) is not 0.

The only significance of x=-4/3 is that it is the point where the function intersects the x axis, or i.e., h(x)=0

Update: oops, I think I miscalculated!
YES IT IS A CRITICAL POINT,
h(x) = (3x+4)^(5/4)
h'(x)= (5/4) (3x+4)^(1/4) (3)
So h'(-4/3) = 0

Answer 3

Simply diffentiate h(x)...

5/4(3x+4)^1/4 (3)=

15/4(3x+4)^1/4=

15
----
(3x+4)^1/4

plug in -4/3

15/{3(-4/3)+4)}^1/4

15/0

Impossible, so no it's not

Answer 4

yes because h'(-4/3) = 0.

h' (x) = 3*(5/4) * (3x + 4) ^ (1/4) = (15/4) * (3x + 4)^(1/4)

Answer 5

yes, just check by plugging -4/3 in... h(x)=0^5/4 = 0 so yes, its a zero or critical pt

Answer 6

Nope

Answer 7

yep