Use the limit definition of the derivative to find f ‘(a) when f(x)=2x^3. What’s the equation of the tangent line at x=1?

Answer 1

f(x) = 2x^3

f(x+h) = 2(x+h)^3 = 2(x^3 + 3x^2h + 3xh^2 + h^3) = 2x^3 +6x^2h + 6xh^2 + 2h^3

f(x+h) - f(x) = 6x^2h + 6xh^2 + 2h^3

[f(x+h)-f(x)]/h = 6x^2 + 6xh + 2h^2

as h approaches zero, [f(x+h)-f(x)]/h approaches 6x^2 + 6x(0)+2(0)^2 = 6x^2

the derivative of 2x^3 = 6x^2

tangent line will have slope = 6x^2. at x=1, 6(1)^2 = 6 = slope of tangent line

it goes through the point (1,2)

y-2 = 6(x-1)

y=6x-6+2 = 6x -4

the equation of the tangent line is y=6x-4

Answer 2

f '(a) = lim (f(a+h) - f(a))/h

h→0

= lim (2a³+4a²h+2ah+2a²+4ah²+2h² -2a³)/h

h→0

= lim 4a²+2ah+2a²+4ah²+2h² = 4a² + 2a²

h→0

f'(a) = 6a²

f(1)= 2

f'(1) = 6

y - 2 = 6(x-1) = 6x-6

y = 6x-4

Answer 3

f^'(a) = lim_x→a(f(x) - f(a)) / (x-a) = lim_x→a(2x³ - 2a³) / (x-a) = lim_x→a(2(x-a)(x²+ax+a²) / (x-a) = lim_x→a2(x²+ax+a²) = 6a²

f(1) = 2 and m_tan = f^'(1) = 6. Eqn of tan line: y - 2 = 6(x - 1)