Question

What is the equation of the line tangent to `f(x)=(2x^2 - 1) / x` at `x=1`?

Answer 1

y = 3x -2

Explanation:

Find the equation in slope-intercept form y = mx + c ,where m represents the gradient and c , the y-intercept. We require to find m and c.
The value of f'(1) is the gradient of the tangent m , and f(1) will enable us to find c.

Begin by simplifying f(x). Divide terms on numerator by x.

`f(x)=(2x^2)/x-1/x=2x-x^-1`

now differentiate using the`color(blue)" power rule"`

`rArrf'(x)=2+x^-2=2+1/x^2`

and`f'(1)=2+1/(1)^2=3=m" (gradient of tangent)"`

hence partial equation is : y = 3x + c

now `f(1)=(2(1)^2-1)/1=1rArr" (1,1) is point on tangent"`

substitute x = 1 , y = 1 into partial equation to find c.

1 = 3(1) + c → c = -2

and equation of tangent at x = 1 is y = 3x -2
graph{(y-((2x^2-1)/x))(y-3x+2)=0 [-10, 10, -5, 5]}