help plz

known function f: R.. R defined by $$f(x)= ax ^{3} + bx ^{2} + cx+d$$ is that f (0) = 4 and its graph has an inflection point at (1,2). Knowing well that the tangent line to the graph of f at the point of abscissa x = 0 in horizontal. Calula a, b, c, and d

Question

help plz

known function f: R..> R defined by $$f(x)= ax ^{3} + bx ^{2} + cx+d$$ is that f (0) = 4 and its graph has an inflection point at (1,2). Knowing well that the tangent line to the graph of f at the point of abscissa x = 0 in horizontal. Calula a, b​​, c, and d

hartnn · Answer

ok, f(0)=4
from the equation, f(0)=... ?
put x=0

anonymous · Answer

$$f(0) =4 ..> a*0^{3} + b*0^{2} c*0 +d =4 ...> d=4$$

$$f''(1)=2 ..> 6a*1+2b=2..> 6a+2b=2.$$

$$f'(0)=0 ..> 3*a*0^{2} +2b*0+c=0 ..> c=0$$


is is correct?

hartnn · Answer

yes.

anonymous · Answer

how to get b

anonymous · Answer

b=0 and a =1/3 ???

anonymous · Answer

At an inflection point, the second derivative is 0.  So f(1)=2, but f''(1)=0.  You should have 6a=-2b, not 6a=-2b+2.

Then at (1,2) you can say$$2=a \times 1^{3}+b \times 1^{2}+c \times x+d=a+b+c+d=a+b$$$$2=a+b=a-3a$$$$a=-1$$$$b=3$$

anonymous · Answer

c and d??

anonymous · Answer

@Aylin

anonymous · Answer

Didn't you already say that c=d=0?

anonymous · Answer

d=4 and c=0

anonymous · Answer

Ah, right.

Then 2=a+b+c+d
2=a+b+d
2=a-3a+4
-2=-2a
a=1
b=-3
c=0
d=4

anonymous · Answer

$$f''(1)=2..> 6ax+2b=2...> 6a+2b=2 ...>  6a+2*0=2 ..> a=1/3$$
$$f''(0)=0 ..> 6a*0+2b=0 ..> 2b=0..> b=0$$

i get a=1/3 and b =0

hartnn · Answer

f"(0)=0 is not correct!

anonymous · Answer

then

anonymous · Answer

The second derivative at (1,2) is equal to 0.

The second derivative at (0,0) is not equal to 0.

anonymous · Answer

then f(0)= ??

anonymous · Answer

f(0)=4

anonymous · Answer

how you get b=3 i dont understand

anonymous · Answer

f''(1) should equal 0 since its an inflection point, not 2.
f(1) = 2.

anonymous · Answer

$$f''(1)=2 ..> 6a+2b=2$$

then ??

anonymous · Answer

No!

$$f''(1)=0$$

anonymous · Answer

Since x = 1 is an inflection point.

anonymous · Answer

f''(1) =0 
f''(2)=0

anonymous · Answer

$$f''(1)=0 ..> 6a+2b=0$$
$$f''(2)=0 ..> 12a+2b=0$$

anonymous · Answer

You use f(0)=4 to find d.  d=4

You use f'(0)=0 to find c.  c=0

You use f''(1)=0 to find b in terms of a.  You get 0=6a+2b -> b=-3a

You use f(1)=2 to find out what a is.  You get$$a+b+c+d=2$$$$c=0,d=4,b=-3a$$$$a-3a+4=2$$$$-2a=-2$$$$a=1$$Then you use b=-3a.  a is 1, so b=-3

That is the entire problem.

anonymous · Answer

I don't know why you're trying to use f''(2)=0, since that isn't true.

anonymous · Answer

because (1,2) is inflection point

anonymous · Answer

Right, but that doesn't have ANYTHING to do with f''(2).

What that means is that f''(1)=0

anonymous · Answer

ok