last set for my test review

Question

Answer 1

12)

The equation y = (1/4)x² and the x-coordinate of the point is - 4.

Find the y-coordinate of the point.

y = (1/4)(-4)² = 4.

Differentiate the equation with respect to x .

y' =(1/4)2x = x/2

Find slope of tangent line at the point P(-4, 4).

m = -4/2 = -2

Point-slope form of line equation : y - y₁ = m(x - x₁), where m = slope and (x₁, y₁) = point.

y - (4) = -2(x - (-4))

y - 4 = -2(x + 4)

y -4 = -2x - 8.

y = -2x - 8+4

y = -2x - 4

The tangent line equation is y = -2x - 4.

Comments

when you deferninate to x x^2 becomes 2x? because of x times x and there is 2x so 2x?

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Derivative of  x^2 is  2 x^(2-1) = 2x

Answer 2

13)

Evaluate the limit.

Answer 3

The function .

Apply derivative on each side.

            (Apply Differentiation rule )

Substitute f(5) = 7 ; f'(5) = 5 : g(5) = 6 ; g'(5) = -1

Therefore .

Answer 4

(11)

a) 

Now simplify the above equation .

Answer 5

(11)

b) 

Substitute x = 3 in the above limit .

Apply L -hospital rule .

Comments

i don't know the ln rule, and idk how the x become an ln 3, and idk how ln3 = 1.0986

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Type ln 3 in your calculator then you will get ln 3 = 1.0986 .