F(x)=4- 2x - x^3
Show that the graph of y=f(x) has no turning points.

It would be F(0)=4

Step-by-step explanation:

Step-by-step explanation:

for a turning point the first derivative of the function must have zero points (= there must be some values of x that make the derivative function to have 0 as result).

the first derivative is the function that calculates the slope of the tangent at every x. so, if that function result is 0, it means the tangent slope is 0 at this point (a horizontal line), and this could be a real candidate for a turning point.

and then, left and right of such a point, the slope of the tangent must have different signs (+ to - or the other way around).

but if there is no real zero point of the first derivative, then there is no turning point anyway.

the first derivative f'(x) = -2 - 3x²

so,

-2 - 3x² = 0

-3x² = 2

x² = -2/3

x = sqrt(-2/3)

now, the square root of a negative number has no real number solution. so, on our grid of real number coordinates there is no point with a horizontal tangent (with the slope of the tangent = f'(x) = 0).

therefore, there cannot be any turning points.

i can't see the picture : /

which expression represents the approximate length of bc?

step-by-step explanation: