Find an equation of the line that is tangent to the graph of f and parallel to the given line

Question

Find an equation of the line that is tangent to the graph of f and parallel to the given line:

Function: f(x) = x^3 +2 Given Line: 3x-y-4 = 0

Answer 1

Step 1: Determine the slope of the given line.

3x − y − 4 = 0 is the same equation as y = 3x - 4, where slope m = 3.

Step 2: Find the derivative of the function f(x).

Derivative of f(x) = x³ + 2 is f'(x) = 3x²

Step 3: Set the derivative equal to slope and solve for x.

Since the derivative represents the slope, we set f'(x) = 3x² = 3. Solving for x gives us x₁ = 1 and x₂ = -1

Step 4: Find the points on the graph of f(x).

For x₁ = 1 we have

f(1) = 1³ + 2 = 3

Thus, the point is (1,3).

For x = −1:

f(−1) = (−1)³ + 2 = 1

Thus, the point is (−1,1).

Step 6: Write the equation of the tangent lines.

For the point (−1,1), we use the formula y - y₀ = m(x - x₀). Hence,

y - 1 = 3(x + 1) which is the same as y - 1 = 3x + 3 equivalent to y = 3x + 2

For the point (1,3), we use the formula y - y₀ = m(x - x₀). Hence,

y - 3 = 3(x - 1) which is the same as y - 3 = 3x - 3 equivalent to y = 3x

Answer:

y = 3x + 2 and y = 3x