determine the point of intersection of the tangents at the points of inflection to the curve f(x)= x^4 – 24x^2 – 2
y' = 4x^3 - 48x
y '' = 12x^2 - 48
at points of inflection, y'' = 0
12x^2 = 48
x^2 = 4
x = ±2
when x=2 , f(2) = 16 - 96 - 2 = -82 , slope = 32 - 96 = -64
when x = -2, f(-2) = -82 , slope = -32 + 96 = 64
1st tangent: slope = -64, point is (2,-82)
-82 = -64(2) + b
b=46
first tangent equation: y = -64x + 46
2nd tangent: slope = 64 , point is (-2,-82)
-82 = 64(-2) + b
b = 46
second tangent is y = 64x + 46
intersection of y = 64x + 46 and y = -64x + 46
64x + 46 = -64x + 46
128x = 0
x = 0, then y = 46
the two tangents intersect at (0,46)
To find the points of inflection on the curve, we need to determine where the second derivative of the function f(x) changes sign. Let's start by finding the second derivative.
Step 1: Find the first derivative:
f(x) = x^4 - 24x^2 - 2
f'(x) = 4x^3 - 48x
Step 2: Find the second derivative:
f''(x) = (4x^3 - 48x)' = 12x^2 - 48
Step 3: Set the second derivative equal to zero to find the critical values:
12x^2 - 48 = 0
Step 4: Solve for x:
12x^2 = 48
x^2 = 4
x = ±2
So, we have two critical points at x = -2 and x = 2.
Step 5: Substitute these critical values back into the original function to find the corresponding y-values:
For x = -2:
f(-2) = (-2)^4 - 24(-2)^2 - 2 = 16 - 96 - 2 = -82
For x = 2:
f(2) = (2)^4 - 24(2)^2 - 2 = 16 - 96 - 2 = -82
The points of inflection are (-2, -82) and (2, -82).
Step 6: Determine the equations of the tangent lines to the curve at the points of inflection.
To find the equations of the tangent lines, we need the slope and a point on each line. The slope of the tangent line at a point is equal to the derivative of the function at that point.
For the point (-2, -82):
The derivative at x = -2 is f'(-2) = 4(-2)^3 - 48(-2) = -32
So, the slope of the tangent line at (-2, -82) is -32.
Using the point-slope form of a linear equation, the equation of the tangent line is:
y - (-82) = -32(x - (-2))
y + 82 = -32(x + 2)
y + 82 = -32x - 64
y = -32x - 146
For the point (2, -82):
The derivative at x = 2 is f'(2) = 4(2)^3 - 48(2) = 32
So, the slope of the tangent line at (2, -82) is 32.
Using the point-slope form of a linear equation, the equation of the tangent line is:
y - (-82) = 32(x - 2)
y + 82 = 32(x - 2)
y + 82 = 32x - 64
y = 32x - 146
Step 7: Find the point of intersection of the two tangent lines.
To find the point of intersection, we need to solve the two equations:
y = -32x - 146
y = 32x - 146
Setting them equal to each other:
-32x - 146 = 32x - 146
Simplifying:
-32x - 32x = -146 + 146
-64x = 0
Dividing by -64:
x = 0
Substituting x = 0 back into either one of the equations, we find the corresponding y-value:
For y = -32x - 146:
y = -32(0) - 146
y = -146
Therefore, the point of intersection of the tangents at the points of inflection is (0, -146).
To determine the point of intersection of the tangents at the points of inflection of a curve, we need to follow these steps:
Step 1: Find the first derivative of the function f(x).
Step 2: Set the first derivative equal to zero and solve for x to find the critical points.
Step 3: Find the second derivative of the function f(x).
Step 4: Set the second derivative equal to zero and solve for x to find the points of inflection.
Step 5: Find the y-coordinate for each point of inflection by substituting the x-coordinate into the original function f(x).
Step 6: Use the first derivative and the points of inflection to find the slope of the tangents at these points.
Step 7: Use the slope from step 6 and the corresponding points of inflection to find the equations of the tangent lines.
Step 8: Find the point of intersection by solving the system of equations formed by the tangent lines.
Let's go through each step in detail for the given curve f(x) = x^4 – 24x^2 – 2.
Step 1: Find the first derivative of the function f(x).
Differentiating f(x) with respect to x, we get:
f'(x) = 4x^3 - 48x
Step 2: Set the first derivative equal to zero and solve for x to find the critical points.
Setting f'(x) = 0:
4x^3 - 48x = 0
Simplifying, we get:
x(x^2 - 12) = 0
This equation is satisfied when x = 0 or x = ±√12
Step 3: Find the second derivative of the function f(x).
Differentiating f'(x) with respect to x, we get:
f''(x) = 12x^2 - 48
Step 4: Set the second derivative equal to zero and solve for x to find the points of inflection.
Setting f''(x) = 0:
12x^2 - 48 = 0
Dividing by 12, we get:
x^2 - 4 = 0
This equation is satisfied when x = ±2
Step 5: Find the y-coordinate for each point of inflection by substituting the x-coordinate into the original function f(x).
For x = 2:
f(2) = 2^4 - 24(2^2) - 2
f(2) = 16 - 96 - 2
f(2) = -82
For x = -2:
f(-2) = (-2)^4 - 24((-2)^2) - 2
f(-2) = 16 - 96 - 2
f(-2) = -82
So we have two points of inflection: (2, -82) and (-2, -82).
Step 6: Use the first derivative and the points of inflection to find the slope of the tangents at these points.
For x = 2:
Slope of the tangent = f'(2) = 4(2)^3 - 48(2)
Slope of the tangent = 32 - 96
Slope of the tangent = -64
For x = -2:
Slope of the tangent = f'(-2) = 4(-2)^3 - 48(-2)
Slope of the tangent = -32 - 96
Slope of the tangent = -128
So the slopes of the tangents at the points of inflection are -64 and -128.
Step 7: Use the slope from step 6 and the corresponding points of inflection to find the equations of the tangent lines.
Using the point-slope form of a line, the equations of the tangent lines are:
For the tangent at (2, -82):
y - (-82) = -64(x - 2)
y + 82 = -64x + 128
y = -64x + 46
For the tangent at (-2, -82):
y - (-82) = -128(x - (-2))
y + 82 = -128x + 256
y = -128x + 174
Step 8: Find the point of intersection by solving the system of equations formed by the tangent lines.
Setting the two equations equal to each other:
-64x + 46 = -128x + 174
64x = 128x - 128
64x - 128x = -128
-64x = -128
x = 2
Substituting the value of x into one of the equations:
y = -64(2) + 46
y = -128 + 46
y = -82
So the point of intersection of the tangents at the points of inflection is (2, -82).