MHB *find coefficients of cubic f(x) given min max & inflexion pt

[karush]

find a, b, c, and d, such the cubic $f(x)=ax^3+bx^2+cx+d$ satisfies the given conditions
Relative maximum (3,3) Relative minimum (5,1) Inflection point (4,2)

I approached this by using the f'(x)= a(3)(x^2)+b(2)(x)+c with the min and max
and f''(x)=6x+2b for inflection pt to get

$27a +6b + c =3$
$75a+10b+c=1$
$24a + 2b = 2$

Then I ran it thru a system equation solver but did not get the coefficeints needed that are in the answer which is ?

$\frac{1}{2}x^3-6x^2+\frac{45}{2}x-24$

probably don't have the dx set up right
Thank ahead for help

 

[tkhunny]

You have not understood the given information. Please rethink. The most common error I have encountered is the failure to realize that a min, max, or inflection point is not JUST a hint about the various derivatives. It is also a hint about the function itself.

Relative maximum (3,3)

f(3) = 3
f'(3) = 0

Relative minimum (5,1)

f(5) = 1
f'(5) = 0

Inflection point (4,2)

f(4) = 2
f"(4) = 0

 

[karush]

ok I see m=0 and POI=0

but in the eq still do not have the right numbers plus in the dx the d is gone

$f'(x)= a(3)(x^2) + b(2)(x) + c and f''(x) a(6)x + b(2)$
min/max
at $(3,3) 27a +6b+c =0$
at $(5,1) 75a+10b+c=0$
inflection pts
at $(4,2) 24a+2b=0$

I don't see how else to get the coefficients without $f'(x)$ and $f''(x)$

 

[MarkFL]

We are given the cubic:

$\displaystyle f(x)=ax^3+bx^2+cx+d$

We are told it passes through the points (3,3), (5,1), (4,2) giving us:

(1) $\displaystyle 27a+9b+3c+d=3$

(2) $\displaystyle 125a+25b+5c+d=1$

(3) $\displaystyle 64a+16b+4c+d=2$

Computing the first derivative, we find:

$\displaystyle f'(x)=3ax^2+2bx+c$

We are given:

(4) $\displaystyle f'(3)=27a+6b+c=0$

(5) $\displaystyle f'(5)=75a+10b+c=0$

Computing the second derivative, we find:

$\displaystyle f''(x)=6ax+2b$

We are given:

(6) $\displaystyle f''(4)=24a+2b=0$

From (6) we find:

$\displaystyle b=-12a$

Substituting for b into either (4) or (5) we find:

$\displaystyle c=45a$

Substituting for b and c into (1) and (3) we find:

$\displaystyle 27a+9(-12a)+3(45a)+d=3$

$\displaystyle 64a+16(-12a)+4(45a)+d=2$

which simplify to:

$\displaystyle 54a+d=3$

$\displaystyle 52a+d=2$

Subtracting, we find:

$\displaystyle 2a=1$

$\displaystyle a=\frac{1}{2}$

and so:

$\displaystyle d=2-26=-24$

$\displaystyle b=-6$

$\displaystyle c=\frac{45}{2}$

 

[karush]

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