How do you find the equation of a line tangent to the function $y = x^{3} - 5 x$ $y=x^3-5x$ at x=1?

Question

5974 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-15T04:38:40

$y = - 2 x - 2$ $y=-2x-2$

Given -

$y = x^{3} - 5 x$ $y=x^3-5x$

Slope of the curve at any point on the curve is given by its
first derivative

$\frac{d y}{d x} = 3 x^{2} - 5$ $dy/dx=3x^2-5$

Slope of the curve exactly at $x = 1$ $x=1$

Substitute $x = 1$ $x=1$ in the first derivative.

$\frac{d y}{d x} = 3 (1^{2}) - 5 = 3 - 5 = - 2$ $dy/dx=3(1^2)-5=3-5 = -2$

The slope of the tangent is $m = - 2$ $m=-2$

The tangent is passing through the point $x = 1$ $x=1$

To find the equation of the tangent , we must know the
y-coordinate at point $x = 1$ $x=1$

For this substitute $x = 1$ $x=1$ in the given function

$y = 1^{3} - 5 (1) = 1 - 5 = - 4$ $y=1^3-5(1)=1-5=-4$

Point $(1, - 4)$ $(1, -4)$ is on the line as well as on the tangent.

We know the slope of the tangent $(m = - 2)$ $(m=-2)$ and the point
through which it passes $(1, - 4)$ $(1, -4)$

$m x + c = y$ $mx+c=y$

$(- 2) (1) + c = - 4$ $(-2)(1)+c=-4$
$c = - 4 + 2 = - 2$ $c=-4+2=-2$

Now we have Y- intercept $c = - 2$ $c=-2$ and slope $m = - 2$ $m=-2$

The equation of the tangent is -

$y = m x + c$ $y=mx+c$

$y = - 2 x - 2$ $y=-2x-2$

Look at the graph

How do you find the equation of a line tangent to the function y=x3−5xy=x^3-5x at x=1?