To find the equation of the tangent to f at x = -1, we need to use the slope-intercept form of a line.
(a) The differential equation provided gives us the slope of the tangent line at any point (x, y) on the curve f. So, to find the slope of the tangent line at x = -1, we substitute x = -1 into the given differential equation:
dy/dx = 10/(x^2+1)
dy/dx = 10/((-1)^2+1)
dy/dx = 10/2
dy/dx = 5
Therefore, the slope of the tangent line at x = -1 is 5. Now we can use the point-slope form of a line to write the equation of the tangent line. Given the point (-1, 8) and the slope m = 5, the equation of the tangent line is:
y - 8 = 5(x + 1)
So, you correctly found the equation of the tangent line for part (a).
Now, let's move on to part (b).
(b) To estimate f(0) using the equation of the tangent line at x = -1, we substitute x = 0 into the equation of the tangent line:
y - 8 = 5(0 + 1)
y - 8 = 5
y = 5 + 8
y = 13
So, you correctly calculated f(0) as 13 for part (b).
Finally, for part (c):
(c) We are given that the integral from -1 to 0 of 10/(x^2 + 1) is approximately 7.854. This integral represents the area under the curve f(x) from x = -1 to x = 0.
By using the Fundamental Theorem of Calculus, we know that the integral from a to b of f(x)dx = F(b) - F(a), where F(x) is the antiderivative of f(x).
Since we have the integral value and we know the antiderivative of 10/(x^2 + 1) is arctan(x), we can write:
F(0) - F(-1) = 7.854
So, by plugging in the values, we get:
arctan(0) - arctan(-1) = 7.854
0 - (-Ï€/4) = 7.854
Ï€/4 = 7.854
Therefore, the integral from -1 to 0 of f(x) is π/4, which equals f(0) - f(-1). Since we know f(-1) = 8 (from the given point), we can solve for f(0):
f(0) - 8 = π/4
f(0) = 8 + π/4
f(0) ≈ 8.785
So, the estimated value of f(0) using integral calculations is approximately 8.785, not 13.
It seems there was an error in part (b) where you found f(0) = 13. Instead, the correct estimate for f(0) should be approximately 8.785 as calculated in part (c).
Keep in mind that these are approximations, and the actual values may differ depending on the level of accuracy required and the numerical methods used for calculation.