Question from 2007

[Teen]

Let:f:R-->R bea differentiable function such at
F'(3)=0
F'(x)<0 when x<3 and when x>3
When x=3,the graph of f has a........

Can some pls explain hoe to draw a graph from that statement...

[BubbleWrapMan]

Think of something like -(x-3)^3, its gradient is always negative except at one point (x=3). Hence it has a stationary point of inflection at x = 3.

[Teen]

I don't understand the third line in the question... Climb too high can u pls explain further .... Thanks man

[aishuwa1995]

Let:f:R-->R bea differentiable function such at
F'(3)=0
F'(x)<0 when x<3 and when x>3
When x=3,the graph of f has a........

Can some pls explain hoe to draw a graph from that statement...

The gradient of the function is negative for all values less than 3 and for all values of x greater than 3 but it has a stationary point at x=3 therefore it must be a stationary point of inflection. As ClimbTooHigh said -(x-3)^3 is the equation of the graph.

[Teen]

Yes thanks guys finally

[BubbleWrapMan]

It's not necessarily that equation, there are other examples, just saying

[aishuwa1995]

It's not necessarily that equation, there are other examples, just saying

haha yeah but it made sense and it was first one I could think of too

[BubbleWrapMan]

Great minds, etc