Question

Складіть рівняння дотичної до графіка f(x)=x^3+1/3 , у точці х0=1

Answer 1

Equation of the Tangent Line to the Graph of f(x) = x^3 + 1/3 at x0 = 1

To find the equation of the tangent line to the graph of the function f(x) = x^3 + 1/3 at the point x0 = 1, we need to determine the slope of the tangent line and the coordinates of the point of tangency.

Step 1: Find the slope of the tangent line The slope of the tangent line to a curve at a given point can be found by taking the derivative of the function and evaluating it at that point.

Let's find the derivative of f(x) = x^3 + 1/3: f'(x) = 3x^2

Now, we can evaluate the derivative at x = 1 to find the slope of the tangent line: f'(1) = 3(1)^2 = 3

Therefore, the slope of the tangent line to the graph of f(x) = x^3 + 1/3 at x0 = 1 is 3.

Step 2: Find the coordinates of the point of tangency To find the y-coordinate of the point of tangency, we substitute x = 1 into the original function: f(1) = (1)^3 + 1/3 = 1 + 1/3 = 4/3

Therefore, the point of tangency is (1, 4/3).

Step 3: Write the equation of the tangent line We have the slope of the tangent line (m = 3) and a point on the line (1, 4/3). We can use the point-slope form of a linear equation to write the equation of the tangent line:

y - y1 = m(x - x1)

Substituting the values, we get: y - 4/3 = 3(x - 1)

Simplifying the equation: y - 4/3 = 3x - 3 y = 3x - 3 + 4/3 y = 3x - 5/3

Therefore, the equation of the tangent line to the graph of f(x) = x^3 + 1/3 at x0 = 1 is y = 3x - 5/3.

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