Determine a,b,c, and d so that the graph of ax^3+bx^2+cx+d has point of inflection (0,0) and rel max (2,4)?

Answer 1

Let f(x) = ax³ + bx² + cx + d. Then we have:
f'(x) = 3ax² + 2bx + c
f''(x) = 6ax + 2b

Now, we know that f passes through the point (0, 0), so f(0) = d = 0. Next, we know this is a point of inflection, so the second derivative changes sign here. This means that f''(0) = 2b = 0, so b = 0. Third, we know that f passes through (2, 4) and that its derivative is zero there, so we have:

f(2) = 8a + 4b + 2c + d = 8a + 2c = 4
f'(2) = 12a + 4b + c = 12a + c = 0

Solving this system of equations:

8a + 2c - 2(12a + c) = 4 - 0
-16a = 4
a = -1/4

12(-1/4) + c = 0
c = 3

So we have f(x) = -1/4 x³ + 3x. And we are done.

Answer 2

f(x)=ax³+bx²+cx+d continous and der. in |R
f'(x)=3ax²+2bx+c with f'(2)=0, f''(2)<0 and f(2)=4
f''(x)=6ax+2b with f''(0)=0 and f(0)=0

--> d=0 and b=0 --> f(x)= a·x³ +cx

12a+c=0 and 8a+2c=4 --> 12a+c=0 and 4a+c=2

--> 1ª-2ª: a= -1/4 and c=3

--> f(x)= -1/4 · (x³-12x)

saludos.