HELP!! a question asks me to show: if f(x) = x^4 then f''(0)=0?

Question

BUT (0.0) is not an inflection point on graph f.

Answer 1

Ah yes, a very important concept in calculus... Just because the second derivative is 0, the point at which it equals 0 is not necessarily an inflection point!

f(x) = x^4
f'(x) = 4x^3
f''(x) =12x^2

12x^2 = 0
x = 0

x <----- 0 ----->
f''.... + ...... +

So f'' is positive to the left and right of x=0.
Therefore f is concave up to the left and right of x=0. So f is not an inflection point.


EDIT: Believe it or not, ALL of the other answers so far are incorrect. I really thought other people would know about this.

Answer 2

Yes it is. (more accurately, at x=0, it is the product of a first degree line with a third degree inflection point.

You'd think that just because the values are positive just before and just after x=0, that this would be an ordinary minimum on f(x). But it is not. The curve actually flattens out.

Think of it as the curve g(x) = x multiplied by the curve h(x) = x^3

h(x) does have an inflection point at x=0. This means that it "flattens out" very close to x=0.

Multiplying it by x will enhance this flattening, not remove it.

The graph will look like a U that has an infininesimally small 'flat bottom'.

Answer 3

f(x)=0
x^4=0
x=0
then f(0)=0
then you have to plug in x=0
then you get (0,0)
it's the origin, or the places where the x and y axes intersect

Answer 4

you plug is zero for x, so 0 to the fourth power is 0...i think.

Answer 5

yes but it must be a point on the derivative of the tangent line (aka 2nd derivative)

Answer 6

if f(x) = x^4 then f''(0)=0
f'(x) = 4x^3

f''(x) =12x^2
f"(0) =12(0) =0