Calculus Derivative - Finding unknown constants

Question

Determine the constants $a$, $b$, $c$, and $d$ so that the curve defined by $y = ax^3 + bx^2 + cx + d$ has a local maximum at the point $(2, 4)$ and a point of inflection at the origin. Sketch the curve. I know to create different equations and use substitution to find the unknown values, but am unsure on how to get the equations.

Answer 1

Let C be the curve in question. C has a local max at (2, 4) if y'(2) = 0 :

First y(2) = 4 ==> 8a + 4b + 2c + d = 4. Second y'(2) = 0 ==> 12a + 4b + c = 0. C has an

inflection point at (0,0) means: y''(0) = 0 ==> 2b = 0. So b = 0. Also (0,0) is on C so:

y(0) = 0 ==> d = 0. So: 8a + 2c = 4 and 12a + c = 0 ==> a = -1/4 and c = 3. So:

(a, b, c, d ) = (-1/4, 0, 3, 0).

Answer 2

The fact that $(2,4)$ and $(0,0)$ lies on the curve gives you your first two equations:

$$y(2) = 4,$$ $$y(0) = 0$$

A local maximum point at $x = 2$ gives you your third equation: $$y'(2) = 0$$

A point of inflection at $x = 0$ gives you your fourth:

$$y''(0) = 0$$

Substitute in to set up a system of $4$ linear equations in $a,b,c,d$.

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There are some observations you can make to simplify things.

Notice that $y(0) = 0 \implies a(0)^3 + b(0)^2 + c(0) + d = 0 \implies d = 0$.

Notice also that $y''(0) = 0 \implies 6a(0) + 2b = 0 \implies b = 0$. This leaves us with:

$$y = ax^3 + cx$$

which can be more easily solved with $y(2) = 4$ and $y'(2) = 0$.

Answer 3

First, differentiate: $$y' = 3ax^2+2bx+c$$ Since we want a local maximum at $(2, 4)$, set the expression for the derivative equal to $0$ and substitute $2$ in for $x$. $$0=12a+4b+c$$ Now, an inflection point at $(0,0$). Differentiate again: $$y''=6ax+2b$$ Since we want an inflection point, set this expression equal to $0$ and substitute $0$ in for $x$: $$0=2b$$ Also, remember that we want $y(2)=4$ and $y(0)=0$. So, we have a system of equations:

• $4=a(2)^3+b(2)^2+c(2)+d$
• $0=a(0)^3+b(0)^2+c(0)+d$
• $0=12a+4b+c$
• $0=2b$

Now just solve the system.

Answer 4

First of all, since the curve has a point of inflection at the origin, it must pass through $(0,0)$ and so $y(0)=0$, implying $d=0$.

Then, the curve also passes through $(2,4)$, so $y(2)=4$ or $$8a+4b+2c=4.\tag1$$

Well you also want your curve to have a maximum at $(2,4)$. This means that the derivative, or $$y'(x)=3ax^2+2bx+c,$$ must be zero at $x=2$. So, $y'(2)=0$ or $$12a+4b+c=0.\tag2$$

Lastly, you want to have a point of inflection at the origin so the second derivative must have $x=0$ as a root of odd multiplicity. Now, $$y''(x)=6ax+2b$$ and we require $y''(0)=0$, implying $b=0$.

Substituting back into $(1)$ and $(2)$ we have $$4a+c=2\tag{$*$}$$and $$12a+c=0\tag{$**$}.$$This is a simple linear system which you can solve to get $a=-1/4$ and $c=3$.

To sum up, $a=-1/4$, $b=0$, $c=3$, $d=0$ and so the required curve is given by $$y=-\frac14x^3+3x$$ which you can easily sketch since you know the stationary points and can instantly find the roots.