Answer:
x-coordinates of relative extrema = 
x-coordinates of the inflexion points are 0, 1
Step-by-step explanation:
Differentiate with respect to x
Differentiate f'(x) with respect to x
At x = ,
We know that if 
then x = a is a point of minima.
So, 
is a point of minima.
For inflexion points:
Inflexion points are the points at which f''(x) = 0 or f''(x) is not defined.
So, x-coordinates of the inflexion points are 0, 1
Answer:
The equation of the line would be y = -3/2x + 7/2
Step-by-step explanation:
In order to find this, we must first use the slope formula with the two point to find the slope.
m(slope) = (y2 - y1)/(x2 - x1)
m = (5 - -1)/(-1 - 3)
m = 6/-4
m = -3/2
Now that we have the slope, we can use that and either point in point-slope form. Then we solve for y.
y - y1 = m(x - x1)
y + 1 = -3/2(x - 3)
y + 1 = -3/2x + 9/2
y = -3/2x + 7/2
Answer:
Five hundred point two
Step-by-step explanation:
500.2
That's a deimal.
The digit 5 is an hundred. Hence, we have
Five hundred point two
Answer:
is there supposed to be a image there for us to see or no?
Answer:
we have the equation y = (1/2)*x + 4.
now, any equation that passes through the point (4, 6) will intersect this line, so if we have an equation f(x), we must see if:
f(4) = 6.
if f(4) = 6, then f(x) intersects the equation y = (1/2)*x + 4 in the point (4, 6).
If we want to construct f(x), an easy example can be:
f(x) = y = k*x.
such that:
6 = k*4
k = 6/4 = 3/2.
then the function
f(x) = y= (3/2)*x intersects the equation y = (1/2)*x + 4 in the point (4, 6)
