Final answer:
To find the equation of the tangent to the curve at the point (-1, -3), we first need to find the slope of the curve at that point using the derivative. Then, we can use the point-slope form of a line to write the equation of the tangent line. The equation of the tangent to the curve f(x) at the point (-1, -3) is y = (13/2)x + 7/2.
Step-by-step explanation:
To find the equation of the tangent to the curve at the point (-1, -3), we first need to find the slope of the curve at that point. We can do this by taking the derivative of the function f(x) and evaluating it at x = -1. Then, we can use the point-slope form of a line to write the equation of the tangent line. The derivative of f(x) is found by using the product rule and the chain rule.
First, let's find the derivative of f(x):
f'(x) = (x - (1/2)x)(d/dx)(3x + 5x^2) + (3x + 5x^2)(d/dx)(x - (1/2)x)
Using the product rule and the chain rule, we simplify:
f'(x) = (x - (1/2)x)(3 + 10x) + (3x + 5x^2)(1 - (1/2))
f'(x) = (x - (1/2)x)(3 + 10x) + (3x + 5x^2)(1/2)
Next, we evaluate the derivative at x = -1:
f'(-1) = (-1 - (1/2)(-1))(3 + 10(-1)) + (3(-1) + 5(-1)^2)(1/2)
f'(-1) = (-3/2)(-7) + (-3 - 5)(1/2)
f'(-1) = (21/2) + (-8)(1/2)
f'(-1) = 21/2 - 4 = 13/2
So, the slope of the tangent line is 13/2. Now, we can use the point-slope form of a line to write the equation of the tangent:
y - y1 = m(x - x1)
Plugging in (-1, -3) as the point and 13/2 as the slope, we have:
y - (-3) = (13/2)(x - (-1))
y + 3 = (13/2)(x + 1)
Finally, we simplify:
y + 3 = (13/2)x + (13/2)
y = (13/2)x + (13/2) - 3
y = (13/2)x + 13/2 - 6/2
y = (13/2)x + 7/2