Conditions for these tasks?
It's about tasks 15, 16, and 17. Thanks in advance:)
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15)
f(x) = ax3 + bx2 + cx + d
(1) f(-x) = -f(x) –> b = d = 0
(2) The turning point can only be at x=0, from which f'(0)=c=7/4
(3) a follows from f(4) = 3
16)
f(x) = ax3 + bx2 + cx + d
(1) Turning point in origin -> point-symmetrical –> b = d = 0
(2) f'(0) = c = 1
(3) f'(3) = 27a + c = 0, from which a
17)
f(x) = ax3 + bx2 + cx + d
(1) f(0) = 6 = d
(2) g has the gradient -1. f intersects g at right angles -> f'(0) = +1 -> c = 1
(3) g touches f -> f'(6) = 3*62*a + 2*6*b + 1 = -1 (grading g)
(4) f(6) = a*63 + b*62 + 6 + 6 = 0
from (3), (4) follow a,b
Do I need only 3 conditions if a 3rd degree function is point-symmetrical to the origin?
The point symmetry simplifies the function so much that you only need two conditions. Because your variables are only a and c. You only need two equations to determine both unknowns. The third condition is the point symmetry from which b=0 and d=0 follows.
You need a condition per variable. So in the most extreme case four conditions. Article 17
Hi,
First, you "translate" the statements into mathematics and then solve them.
Task 15
A polynomial function of the third degree generally has the following function equation:
f(x) = ax³ + bx² + cx + d
This information already provides us with a lot, namely:
Means: f(4) = 3. This can now be used:
3 = a*4³ + c*4
This means that the first derivative at x = 0 has the solution 7/4, because the first derivative at the point x0 gives the slope of the tangent at this point:
3a*0² + c = 7/4
c = 7/4
You can now insert this into the equation you just set up and calculate a:
3 = 64a + 4*(7/4)
3 = 64a + 7
-4 = 64a
a = -4/64 = -1/16
The function equation is f(x) = -1/16 x³ + 7/4.
Task 16
Again, set up the general equation: f(x) = ax³ + bx² + cx + d
d = 0, since the family passes through the origin. The function is also point-symmetric, so again b = 0:
f(x) = ax³ + cx.
Again: The first derivative at the point x0 corresponds to the slope at the point:
f'(x) = 3ax² + c
The term 3ax² disappears because x = 0, resulting in c = 1.
Peak means: The first derivative of the function has the function value 0 at the point x = 3:
0 = 3a*3² + 1 (we know that c = 1)
0 = 27a + 1
-1 = 27a
a = -1/27
The functional equation is:
f(x) = -1/27 x³ + x.
Task 17
We reformulate this in the well-known form:
y = g(x) = -x + 6.
f(x) = ax³ + bx² + cx + d
From this we get the following information:
The result is: 3a*6² + 2*6*b + c = -1
That is: g is the so-called normal to the tangent at the point x = 0. The tangent has the gradient -1/m1, so the gradient at the point x = 0 is m2 = 1:
3a*0² + 2*0*b + c = 1
This results in c = 1.
We now rearrange the formula 3a*6² + 2*6*b + c = -1 for a or b – it doesn't matter:
108a + 12b + 1 = -1
12b = -108a -2
b = -9a -1/6
Now we know that f(6) = 0:
a*6³ + b*6² + c*6 + d
We know that b = -9a -1/6, c = 1 and d = 6; we replace all of this and solve for a:
216a + 36*(-9a -1/6) + 1*6 + 6 = 0
216a – 324a -6 +12 = 0
-108a = -6
a = 1/18
Now insert this into the equation solved for b:
b = -9a – 1/6 = -9/18 – 3/18 = -12/18 = -2/3
So the equation is:
f(x) = 1/18 x³ – 2/3 x² + x + 6.
If anything is still unclear, please feel free to ask.
LG