HELP PLEASE!!!!!!!!!!!!!!!!! BEEN STUCK ON THIS FOREVER!
On the interval [-6,6], f(x) is continuous and differentiable. If f ‘(x) = (x2 – 4)(x + 1)2, briefly justify the following conclusion:
“f ‘(x) has an x-intercept at x = -1 but x = -1 is not a relative extrema on the graph of f(x) because…”

Question

HELP PLEASE!!!!!!!!!!!!!!!!! BEEN STUCK ON THIS FOREVER!
On the interval [-6,6], f(x) is continuous and differentiable. If f ‘(x) = (x2 – 4)(x + 1)2, briefly justify the following conclusion:
“f ‘(x) has an x-intercept at x = -1 but x = -1 is not a relative extrema on the graph of f(x) because…”

anonymous · Answer

because the derivative does not change sign there

anonymous · Answer

$$f '(x) = (x^2 – 4)(x + 1)^2$$ the $(x+1)^2$ term is always positive unless it is zero
so while the derivative is in fact 0 at $x=-1$ the derivative is negative on both sides, so the function does not change direction

anonymous · Answer

so, you are looking at this function on a graph that shows -6 to 6. the function is continuous on this interval (no holes or jumps). they tell you that x intercepts the y-axis at -1, but this is not one of the extrema.
"extrema" are the maxes and mins of your function, so this is basically saying that even though it is an intercept it is not the lowest or highest point on your graph.

anonymous · Answer

the way you determine if it is a max/min or not is to check the sign of the derivative on either side.  if it were a max/min, then the derivative would be positive on one side and negative on the other.  this is because the derivative is a graph of the slope of the function, and an increasing slope is positive whereas a decreasing slope is -. for something to be a 'peak' it must go from positive to negative slope; and vice versa for a 'valley'

anonymous · Answer

@Gwynnia @satellite73 thanks so much guys!