Answer:
Part I)
[tex]a=-1\text{ and } c=2[/tex]
Giving the equation:
[tex]y=-x^3+x+2[/tex]
Part II)
(-1, 2)
Step-by-step explanation:
We have the equation:
[tex]y=ax^3+x+c[/tex]
Where a and c are constants.
We are given gradient/slope of the curve at the point (1, 2) is -2.
Part I)
We know that the gradient of the curve exactly at the point (1, 2) is -2.
So, we will differentiate our equation. We have:
[tex]\displaystyle y=ax^3+x+c[/tex]
Differentiate both sides with respect to x:
[tex]\displaystyle y^\prime=\frac{d}{dx}\big[ax^3+x+c\big][/tex]
Expand the right:
[tex]\displaystyle y^\prime=\frac{d}{dx}\big[ax^3]+\frac{d}{dx}\big[x\big]+\frac{d}{dx}\big[c\big][/tex]
Since a and c are simply constants, we can move them outside:
[tex]\displaystyle y^\prime=a\frac{d}{dx}\big[x^3]+\frac{d}{dx}\big[x\big]+c\frac{d}{dx}\big[1\big][/tex]
Differentiate. Note that the derivative of a constant is simply 0.
[tex]\displaystyle y^\prime=3ax^2+1[/tex]
We know that at the point (1, 2), the gradient/slope is -2.
Hence, when x is 1 and when y is 2, y’ is -2.
So (we don’t have a y so we can ignore it):
[tex]-2=3a(1)^2+1[/tex]
Solve for a. Simplify:
[tex]-2=3a+1[/tex]
Subtract 1 from both sides yield:
[tex]-3=3a[/tex]
So, it follows that:
[tex]a=-1[/tex]
Therefore, our derivative will be:
[tex]y=-3x^2+1[/tex]
Now, we can go back to our original equation:
[tex]y=ax^3+x+c[/tex]
Since we know that a is -1:
[tex]y=-x^3+x+c[/tex]
Recall that we know that a point on our curve is (1, 2).
So, when x is 1, y is 2. By substitution:
[tex]2=-(1)^3+1+c[/tex]
Solve for c. Simplify:
[tex]2=-1+1+c[/tex]
Then it follows that:
[tex]c=2[/tex]
Hence, our entire equation is:
[tex]y=-x^3+x+2[/tex]
Part II)
Our derivative is:
[tex]y^\prime=-3x^2+1[/tex]
If the gradient is -2, y’ is -2. Hence:
[tex]-2=-3x^2+1[/tex]
Solve for x. Subtracting 1 from both sides yields:
[tex]-3=-3x^2[/tex]
So:
[tex]x^2=1[/tex]
Then it follows that:
[tex]x=\pm1\\[/tex]
Since we already have the point (1, 2) where x is 1, our other point is where x is -1.
Using our equation:
[tex]y=-x^3+x+2[/tex]
We can see that our second point is:
[tex]y=-(-1)^3+(-1)+2[/tex]
Simplify:
[tex]y=1-1+2=2[/tex]
So, our second point where the gradient is -2 is at (-1, 2).
