Question

How do you find a tangent line parallel to secant line?

Answer 1

To find the tangent lines parallel to this secant line, we will take the function's derivative, $f^{'}\left( x \right)$, by considering an example for the function $f\left( x \right)$, solve for x. Such that, the line tangent must be parallel to the secant line passing through the value of x.

Complete step by step answer:
You need to find a tangent line parallel to a secant line using the Mean Value Theorem.
The Mean Value Theorem states that if you have a continuous and differentiable function, then
$f^{'}\left( x \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$
To use this formula, you need a function $f\left( x \right)$. Hence let us use $f\left( x \right) = - {x^3}$as an example.
$f\left( x \right) = - {x^3}$ ……………. 1
And let us consider the value of a and b as: $a = - 2$and $b = 2$ for the interval for the secant line. This is the line that passes through the points $\left( { - 2,8} \right)$ and $\left( {2, - 8} \right)$.
Now, let us substitute the values of the function a and b and we know that the slope of this line will be:
$= \dfrac{{ - 8 - 8}}{{2 - \left( { - 2} \right)}}$
$= \dfrac{{ - 16}}{4} = - 4$
To find the tangent lines parallel to this secant line, from equation 1 we will take the function's derivative, $f^{'}\left( x \right)$and set it equal to -4, and then solve for x i.e.,
$f\left( x \right) = - {x^3}$
$\Rightarrow - 3{x^2} = - 4$ ………………… 2
Hence, solve for x in equation 2 we get:
$x = \pm \sqrt {\dfrac{4}{3}}$
So, the lines tangent to $y = - {x^3}$ at $x = \sqrt {\dfrac{4}{3}}$and $x = - \sqrt {\dfrac{4}{3}}$ must be parallel to the secant line passing through $x = 2$ and $x = - 2$.

Note: We must note that Secant line is one that intersects two points in a line, whereas tangent line intersects exactly one point on the curve. We have used Mean value theorem as it is the relationship between the derivative of a function and increasing or decreasing nature of function. It basically defines the derivative of a differential and continuous function.