Question

Find the tangent line to the following function at the given value. y = (3x^2 + x - 4)(2x^3+ 3x + 1) at x = 1

Answer 1

f(x) = (3·x^2 + x - 4)·(2·x^3 + 3·x + 1)

f(x) = 3·x^2·2·x^3 + 3·x^2·3·x + 3·x^2·1 + x·2·x^3 + x·3·x + x·1 - 4·2·x^3 - 4·3·x - 4·1

f(x) = 6·x^5 + 9·x^3 + 3·x^2 + 2·x^4 + 3·x^2 + x - 8·x^3 - 12·x - 4

f(x) = 6·x^5 + 2·x^4 + x^3 + 6·x^2 - 11·x - 4

f'(x) = 30·x^4 + 8·x^3 + 3·x^2 + 12·x - 11

tangent at x = 1

f(1) = 6·1^5 + 2·1^4 + 1^3 + 6·1^2 - 11·1 - 4 = 0

f'(1) = 30·1^4 + 8·1^3 + 3·1^2 + 12·1 - 11 = 42

t(x) = f'(1)·(x - 1) + f(1)

t(x) = 42·(x - 1) + 0

t(x) = 42·x - 42 + 0

t(x) = 42·x - 42

Answer 2

The tangent line using limits/ derivatives is y=42x-42

Explanation: