Explanation:
The cubic ...
f(x) = ax³ +bx² +cx +d
has derivatives ...
f'(x) = 3ax² +2bx +c
f''(x) = 6ax +2b
<h3>a)</h3>
By definition, there will be a point of inflection where the second derivative is zero (changes sign). The second derivative is a linear equation in x, so can only have one zero. Since it is given that a≠0, we are assured that the line described by f''(x) will cross the x-axis at ...
f''(x) = 0 = 6ax +2b ⇒ x = -b/(3a)
The single point of inflection is at x = -b/(3a).
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<h3>b)</h3>
The cubic will have a local extreme where the first derivative is zero and the second derivative is not zero. These will only occur when the discriminant of the first derivative quadratic is positive. Their location can be found by applying the quadratic formula to the first derivative.

There will be zero or two local extremes. A local extreme cannot occur at the point of inflection, which is where the formula would tell you it is when there is only one.
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<h3>c)</h3>
Part A tells you the point of inflection is at x= -b/(3a).
Part B tells you the midpoint of the local extremes is x = -b/(3a). (This is half the sum of the x-values of the extreme points.) You will notice these are the same point.
The extreme points are located symmetrically about their midpoint, so are located symmetrically about the point of inflection.
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Additional comment
There are other interesting features of cubics with two local extremes. The points where the horizontal tangents meet the graph, together with the point of inflection, have equally-spaced x-coordinates. The point of inflection is the midpoint, both horizontally and vertically, between the local extreme points.
Answer:
Q ' (3,4)
Step-by-step explanation:
When we reflect across the x axis, the y coordinate becomes -y
Q (3,-4) becomes Q' (3,- -4)
Q ' (3,4)
2,000
10,000 divided by 5 = 2,000
Step-by-step explanation:
there are 2 main approaches (in their core they are the same, of course, but they look different) :
1. use one equation to express one variable by the second, and then use this in the second equation to solve for the remaining variable. with that you go back into the first equation and solve for the first variable.
2. multiply both equations by certain factors as needed, then add both results, so that there is one summary equation with only one variable, and then solve for it. then use that in one of the original equations and solve for the second variable.
using 1.
3x + 2y = 1
3x = 1 - 2y
x = (1 - 2y)/3
2x + 7y = 3
2(1 - 2y)/3 + 7y = 3
2(1 - 2y) + 21y = 9
2 - 4y + 21y = 9
17y = 7
y = 7/17
x = (1 - 2y)/3 = (1 - 2×7/17)/3 = (1 - 14/17)/3 =
= (17/17 - 14/17)/3 = 3/17 / 3 = 1/17
using 2.
multiply first equation by 2, the second by -3
6x + 4y = 2
-6x - 21y = -9
----------------------
0 -17y = -7
-17y = -7
17y = 7
y = 7/17
and so on (as under 1.).
Answer:
-14
Step-by-step explanation: