$y$ coordinate of given turning point on a curve is $10$

Question

I have a question that I do not know how to do, all the questions I attempted had $x, y$ values to them. However this is a bit difficult for me, can someone point me in the right direction please:

The gradient of a curve is given by the equation: $$ \frac{dy}{dx} = 6x^2 - 12x.$$ the $y$-coordinate of the turning point on the curve is $10$, $P(x,10)$. Find the equation of the curve.

Any insight would be great.

Answer 1

• You may call $F(x)$ the function corresponding to the curve.

• The gradient here is the derivative of F. So you may write : $F'(x)= f(x)= 6x^2 -12 x$.

FIRST INFORMATION YOU CAN DERIVE REGARDING $F$

• The fact that $f$ is the derivative of $F$ means that $F$ is a primitive of $f$, which simply means that $F$ is a function such that its derivative is $f$ ( not " the function " but " a function", since there are infinitely many functions which have $f$ as derivative). Using your knowledge of derivation rules, you can guess that the primitive $F(x)$ you are looking for is is of the form : $ 2x^{2+1} -6x^{1+1} +C= 2x^3-6x^2 +C$ ( with $C$ a constant).

SECOND INFORMATION YOU CAN DERIVE REGARDING $F$

• What you also know is that when $F(x)$ evaluates at $10$, i. e. when $F(x)=10$ , $F$ is at a minimum , meaning that its derivative evaluates at $0$, i.e. $f(x)=0 $.

• The fact that $F$ is at a minimum also means that the slope of the tangent to its graph is increasing. This amounts to saying that $F''(x)=f'(x)= 12x-12 \gt 0$.

• Denoting by $(a, F(a))= (a, 10) $ the turning point , and adding these informations, we have :

$F(a)=10 \implies f(a)=0\space \&\space f'(a)\gt 0$

• If we are lucky , there is only one number $a$ that satisfies the consequent of the conditional , so this number will also be the only one that satisfies the antecedent.

• Solving for $a$ the equation $F'(a)= f(a)=0$ gives :

$f(a)= 0 \iff 6a^2 -12 a =0 \iff 6a(a-2) =0 \iff a=0 \space\mathbb {OR}\space a=2$

• Solving for $a$ the equation $ F''(a) = f'(a) \gt 0$ gives :

$f'(a) \gt 0 \iff 12x-12 \gt 0 \iff 12(x-1) \gt 0 \iff x \gt 1$.

• The only value that satisfies both equations is $a = 2$ .

• A little reasoning allows us to say that $F(2)=10$ , namely : if $a\neq2$ , the consequent (above) is not satisfied and therefore the antecedent $F(a)=10$ is false; but we know ( by hypothesis) that $F(a)=10$, so $a=2$, implying that the turning point is $(2, F(2))= (2,10)$.

GATHERING THESE $2$ INFORMATIONS

• Finally, we have 2 informations regarding $F$ and these 2 informations allow us to identify this function :

(1°) $F$ is a primitive of $f$, meaning that $ F(x)= 2x^3-6x^2 +C$ ( see above)

(2) $F(a)=F(2)=10 $

• At this point it is a matter of algebra to solve for $C$ in order to get a fully determined expression for $F$.

Note : with (1°) above, you can obtain an expression for $F(2)$ that involves $C$.

https://www.desmos.com/calculator/esn6aon0xe : modify $C$ in order to get the right turning point and then the right constant

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Answer 2

You're given a derivative $\frac{dy}{dx}$, where $y = f(x)$ (i.e. $y$ is a function of $x$). To get $y$, you need to take the indefinite integral of $\frac{dy}{dx}$, i.e. you need to do:

$$\int \frac{dy}{dx} \; dx = \int 6x^2 - 12x \; dx$$

This will give you $y + c$. The $+ \; c$ is called the constant of integration and is discussed here in Paul's Online Math Notes.

You will need to use the extra information (about the turning point) to figure out the proper value of $c$.