
How do you find a tangent line parallel to secant line?
Answer
540.9k+ views
Hint: 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.
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.
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