How do you find a tangent line parallel to secant line?
Answer
557.4k+ 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.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Which among the following are examples of coming together class 11 social science CBSE

