Find a curve y=f(x) with the following properties (i) y"=6x and (ii) its graph passes through the point (1,1) and has a horizontal tangent there.

Question

Find a curve y=f(x) with the following properties (i) y"=6x and (ii) its graph passes through the point (1,1) and has a horizontal tangent there.

Answer 1

y'' = 6x

Integrate:
y' = 3x^2 + c1
Since it was said that it has a horizontal tangent at (1,1), then we let y' (the slope) be equal to 0:
0 = 3x^2 + c1
0 = 3(1)^2 + c1
0 = 3 + c1
c1 = -3

Integrate y' further (and substitute c1 = -3):
y' = 3x^2 - 3
y = x^3 - 3x + c2
It passes through the point (1,1). Thus,
1 = (1)^3 - 3(1) + c2
1 = 1 - 3 + c2
1 = -2 + c2
c2 = 3

Therefore,
y = x^3 - 3x + 3

hope this helps~ `u`

Answer 2

To find a curve that satisfies the given conditions, we need to integrate twice.

Let's start by integrating the given equation (i) with respect to x once. This will give us the equation for the first derivative:

y' = ∫6xdx
= 3x^2 + C1

Here, C1 is the constant of integration.

Since the graph passes through the point (1,1) and has a horizontal tangent there, we can use this information to find the value of the constant C1.

Substituting the coordinates of the point (1,1) into the equation, we get:

1 = 3(1)^2 + C1
1 = 3 + C1
C1 = -2

So, the equation for the first derivative becomes:

y' = 3x^2 - 2

Now, let's integrate the first derivative equation again with respect to x to find the equation for f(x):

y = ∫(3x^2 - 2)dx
= x^3 - 2x + C2

Here, C2 is the constant of integration.

Now we have the equation for the curve, satisfying both conditions (i) and (ii), as:

y = x^3 - 2x + C2

To find the specific value of the constant C2, we can substitute the point (1,1) into the equation:

1 = (1)^3 - 2(1) + C2
1 = 1 - 2 + C2
C2 = 2

Therefore, the curve y = f(x) that satisfies the given conditions is:

y = x^3 - 2x + 2

Answer 3

To find a curve satisfying the given conditions, we need to integrate the second derivative of the curve, y", with respect to x twice. This will give us the original function, y.

(i) First, integrate y" with respect to x once to find y':

∫(y")dx = ∫(6x)dx
y' = 3x^2 + C1

(ii) Next, integrate y' with respect to x once more to find y:

∫(y')dx = ∫(3x^2 + C1)dx
y = x^3 + C1x + C2

Now, we need to find the values of C1 and C2 using the given conditions.

(iii) The graph passes through the point (1,1). Substituting x = 1 and y = 1 into the equation y = x^3 + C1x + C2, we get:

1 = 1^3 + C1(1) + C2
1 = 1 + C1 + C2

This gives us the equation: C1 + C2 = 0

(iv) The graph has a horizontal tangent at (1,1), which means the derivative of y at x = 1 should be 0. Differentiating y = x^3 + C1x + C2 with respect to x, we get:

y' = 3x^2 + C1

When x = 1, y' = 0 implies:

0 = 3(1)^2 + C1
C1 = -3

Now, substituting C1 = -3 into the equation C1 + C2 = 0, we can find C2:

-3 + C2 = 0
C2 = 3

Therefore, the curve that satisfies the given conditions is:

y = x^3 - 3x + 3