Final answer:
The derivative f'(1) is 11. The slope m of the tangent line at (1,6) is 11. The equation of the tangent line to the curve at the point (1,6) is y = 11x - 5.
Step-by-step explanation:
To find f'(1), we need to find the derivative of the function F(x) = 7x² - x³. The derivative represents the rate at which the function is changing at a specific point. In this case, the derivative represents the slope of the tangent line to the curve at x = 1. To find f'(1), we can use the power rule of derivatives. The power rule states that the derivative of x^n is n*x^(n-1). Applying the power rule to each term of F(x), we get: f'(x) = 14x - 3x². Plugging in x = 1, we get f'(1) = 14(1) - 3(1²) = 14 - 3 = 11.
To find the slope m of the tangent line to the curve y = 7x² - x³ at the point (1,6), we can use the derivative f'(x) we found earlier. Plugging in x = 1 into f'(x), we get f'(1) = 11. Therefore, the slope m of the tangent line at (1,6) is 11.
Now we can find an equation of the tangent line to the curve y = 7x² - x³ at the point (1,6). We can use the point-slope form of the equation of a line, which is y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line and m is the slope of the line. Plugging in the values of (1,6) for (x₁, y₁) and 11 for m, we get y - 6 = 11(x - 1). Rearranging the equation to slope-intercept form, we get y = 11x - 5. Therefore, the equation of the tangent line to the curve at the point (1,6) is y = 11x - 5.