Find an equation of the tangent line to the graph of the following function f at the specified point.

Question

f(x)=(x^3+5)(3x^2-4x+2); (1,6)

Answer 1

Given: f(x) = (3x^2 + 5)*(3x^2 - 4x + 2) and Point (1,6)

Find: Equation of tangent line at given point.

Solution:

Use product rule to find f'(x) then f'(1)

f'(x) = [6x]*( 3x^2 - 4x + 2) + (3x^2 + 5)*[6x - 4]

f'(1) = [6]*(3 - 4 + 2) + (3 + 5)*[6 - 4]

f'(1) = (6)*(1) + (8)*(2)

f'(1) = 6 + 16

f'(1) = 22

Slope of tangent line, at given point is 22

Now, find the equation of the tangent line.

y = mx + b

y = (22)x + b

6 = (22)(1) + b Use given point (x,y) = (1, 6)

-16 = b

Equation of tangent line at given point is:

y = (22)x - 16