Determine the equation of the tangent at the point indicated
F(x)= (x²+1)^3 at x=1
To find the equation of the tangent at x=1, we first need to find the slope of the tangent at that point.
We can do this by finding the derivative of the function F(x).
F'(x) = 3(x²+1)² * 2x = 6x(x²+1)²
Now substitute x=1 into this derivative to find the slope at x=1.
F'(1) = 6(1)(1²+1)² = 6(1)(2)² = 6(1)(4) = 24
So the slope of the tangent at x=1 is 24.
Next, we need to find the y-coordinate of the point where the tangent touches the curve. To do this, substitute x=1 into the original function F(x).
F(1) = (1²+1)³ = (2)³ = 8
So the y-coordinate at the point of tangency is 8.
Now we can use the point-slope form of a linear equation to write the equation of the tangent line.
y - y1 = m(x - x1)
Using the point (1, 8) and the slope m = 24, the equation of the tangent line is:
y - 8 = 24(x - 1)
Simplifying the equation gives:
y - 8 = 24x - 24
So the equation of the tangent at x=1 is:
y = 24x - 16
To determine the equation of the tangent at the point indicated, we need to find the slope of the tangent line and the coordinates of the point.
1. Find the derivative of the function f(x) = (x^2 + 1)^3 with respect to x.
f'(x) = 3(x^2 + 1)^2 * 2x = 6x(x^2 + 1)^2.
2. Evaluate the derivative at x = 1 to find the slope of the tangent line.
f'(1) = 6(1)(1^2 + 1)^2 = 6(1)(2)^2 = 24.
3. Substitute x = 1 into the original function to find the y-coordinate of the point.
f(1) = (1^2 + 1)^3 = (2)^3 = 8.
The point at which the tangent touches the curve is (1, 8), and the slope of the tangent line is 24.
Now we can use the point-slope form of a linear equation to find the equation of the tangent line:
y - y1 = m(x - x1),
where (x1, y1) is the point on the line and m is the slope.
Plugging in the values we found:
y - 8 = 24(x - 1).
Simplifying, we get:
y - 8 = 24x - 24.
Finally, rearranging the equation to slope-intercept form, we have:
y = 24x - 16.
Therefore, the equation of the tangent line to the function f(x) = (x^2 + 1)^3 at x = 1 is y = 24x - 16.