Hi there!
a.
At x = 1 for the function 'f', we are given that:
f'(1) = 2
f(1) = -1 ⇒ (1, -1)
We can use the point-slope formula to make a line with the information:
y - y₁ = m(x - x₁)
Plug in the derivative for 'm' and the coordinates:
y - (-1) = 2(x - 1)
y + 1 = 2x - 2
y = 2x - 3
B.
A point of inflection is where the second derivative changes signs, or f''(a) = 0.
According to the information given, f''(1) = 0, so there IS a point of inflection at this point.
C.
Solve for g'(1) using the given equation:
g'(x) = x²[2f(x) + f'(x)]
Use the given information:
g'(1) = (1²)[2f(1) + f'(1)] = (1)(-2 + 2) = 0
The slope is 0, so the graph is a horizontal line.
We are given that g(1) = 3, so:
y = 3 is our equation for the tangent line.
D.
We can use the power rule to solve.
[tex]\frac{dy}{dx}f(x) * g(x) = f'(x)*g(x) + g'(x)* f(x)[/tex]
Let:
[tex]f(x) = x^2\\g(x) = 2f(x) + f'(x)[/tex]
Solve:
[tex]2x(2f(x) + f'(x)) + x^2(2f'(x) + f''(x)) \\ \\=4xf(x) + 2xf'(x) + 2x^2f'(x) + x^2f''(x)\\\\ = 4xf(x) + (2x + 2x^2)f'(x) + x^2f''(x)\\\\= 4xf(x) + 2x(x + 1)f'(x) + x^2f''(x)[/tex]
We can find this expression's equivalent for x = 1, and use this value to determine whether there is a local min or max at this value.
[tex]g''(1) = 4(1)(-1) + 2(1)(2)(2) + (1^2)(0) = -4 + 6 = 2[/tex]
Thus, since the second derivative is positive at this point, there is a LOCAL MIN at x = 1.
