For f(x)=−x3+8x, what is the equation of the line tangent to x=1?

Question

Elodie Williamson · Accepted Answer

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

\frac{d y}{d x} = - 3 x^{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 \cdot {(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 - y_{1}) = m (x - x_{1})

(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

For f(x)=-x^{3}+8x, what is the equation of the line tangent

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