Question

For `f(x) =-x^3+8x`, what is the equation of the line tangent to `x =1`?

Answer 1

`y= 5x -2`

Explanation:

Firstly, you find the slope of the curve, i.e. by derivative,

`dy/dx` = `-3x^2 + 8` is the slope of the tangent.

Now as it is given that you need the find the equation of the tangent at point where `x` = 1,

Slope at `x` =1

= `-3.(1) ^2` + 8
= 5

Now you have got the slope and you need a point where this tangent passes through.
Now see, at `x=1` , the curve and the tangent meet.

So you can replace this `x` in the equation to get the value of `y`.

SO,

y=-1 + 8 = 7

So, the curve and the tangent both pass through ( 1,7).

Now equation of the tangent passing through (1,7 ) and with slope m= 5,

`(y-y1)=m(x-x1)`

`(y-7) = 5(x-1)`

`y-7 = 5x -5`

So,

`y= 5x -2` is the equation go the tangent of the given curve at x= 1