Question

Find the tangent line to f (x) = 15 - 2x^3 at x = 1.

Answer 1

AnswEr :

• The equation of the tangent is 6x + y = 19.

Given,

A tangent joins with a line which has a line equation as follows,

$\sf \: f(x) = 15 - 2 {x}^{3}$

1 is the x-intercept made by the tangent with the line.

For the y-intercept,

$\sf f(1) = 15 - 2( {1}^{3} ) \\ \\ \implies \sf \: f(1) = 13$

Thus,the tangent meets the line at the point (1,13).

Slope of tangent at x = 1 would be :

Firstly, differentiating f(x) w.r.t x on both sides

$\sf \: \dfrac{f(x)}{dx} = \dfrac{d(15 - 2 {x}^{3} )}{dx} \\ \\ \longrightarrow \sf \: f'(x) = - 6 {x}^{2}$

Slope at x = 1,

$\longrightarrow \: \sf \: f'(1) = - 6(1) {}^{2} \\ \\ \longrightarrow \boxed{ \boxed{\sf \: m = - 6}}$

Point-Slope form :

$\sf \: (y - y_1) = m(x - x_1) \\ \\ \implies \sf \: (y - 13) = - 6(x - 1) \\ \\ \implies \sf \: y - 13 = - 6x + 6 \\ \\ \implies \underline{ \boxed{ \sf \: 6x + y - 19 = 0}}$