Let f be a twice differentiable function with f(1) = -1, S'(1) = 2, and f"(1)=0. Let gbe a function whose derivative is given by g'(x) = x? [25 (x)+ f'(x)] for all x. a. Write an equation for the line tangent to the graph of fat x=1.

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.

Let:

Solve:

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.

Thus, since the second derivative is positive at this point, there is a LOCAL MIN at x = 1.

because linear equations are a lot simpler than quadratics equations, and linear equations formula is a lot easier to memorize. that's why i think they're harder.