Find The Equation Of The Line Tangent To The Graph Of F ( X ) = − 5 Csc ⁡ ( X ) − 10 3 3 − 10 Π 9 − 2 F(x)=-5 \csc (x)-\frac{10 \sqrt{3}}{3}-\frac{10 \pi}{9}-2 F ( X ) = − 5 Csc ( X ) − 3 10 3 ​ ​ − 9 10 Π ​ − 2 At X = − Π 3 X=-\frac{\pi}{3} X = − 3 Π ​ . Provide Your Answer Below:

Introduction

In calculus, the concept of a tangent line to a curve is crucial in understanding the behavior of functions. Given a function $f(x)$, the tangent line at a point $x=a$ is a line that just touches the curve at that point and has the same slope as the curve at that point. In this article, we will explore how to find the equation of the tangent line to the graph of $f(x)=-5 \csc (x)-\frac{10 \sqrt{3}}{3}-\frac{10 \pi}{9}-2$ at $x=-\frac{\pi}{3}$.

Understanding the Function

The given function is $f(x)=-5 \csc (x)-\frac{10 \sqrt{3}}{3}-\frac{10 \pi}{9}-2$. To find the equation of the tangent line, we need to first understand the behavior of this function. The function involves the cosecant function, which is the reciprocal of the sine function. The cosecant function has a period of $2\pi$, and its range is all real numbers.

Finding the Derivative

To find the equation of the tangent line, we need to find the derivative of the function. The derivative of a function $f(x)$ is denoted as $f'(x)$ and represents the rate of change of the function with respect to $x$. Using the chain rule and the fact that the derivative of $\csc (x)$ is $-\csc (x) \cot (x)$, we can find the derivative of the given function.

$$f'(x)=5 \csc (x) \cot (x)$$

Evaluating the Derivative at the Given Point

Now that we have the derivative of the function, we need to evaluate it at the given point $x=-\frac{\pi}{3}$. This will give us the slope of the tangent line at that point.

$$f'\left(-\frac{\pi}{3}\right)=5 \csc \left(-\frac{\pi}{3}\right) \cot \left(-\frac{\pi}{3}\right)$$

Using the fact that $\csc \left(-\frac{\pi}{3}\right)=-\sqrt{3}$ and $\cot \left(-\frac{\pi}{3}\right)=-\frac{1}{\sqrt{3}}$, we can simplify the expression.

$$f'\left(-\frac{\pi}{3}\right)=5 \cdot -\sqrt{3} \cdot -\frac{1}{\sqrt{3}}=\frac{5}{1}=5$$

Finding the Equation of the Tangent Line

Now that we have the slope of the tangent line, we can use the point-slope form of a line to find its equation. The point-slope form of a line is given by:

$$y-y_1=m(x-x_1)$$

where $m$ is the slope of the line, and $(x_1,y_1)$ is a point on the line. In this case, the slope of the tangent line is $5$, and the point is $\left(-\frac{\pi}{3},f\left(-\frac{\pi}{3}\right)\right)$.

$$y-f\left(-\frac{\pi}{3}\right)=5\left(x-\left(-\frac{\pi}{3}\right)\right)$$

To find the value of $f\left(-\frac{\pi}{3}\right)$, we need to substitute $x=-\frac{\pi}{3}$ into the original function.

$$f\left(-\frac{\pi}{3}\right)=-5 \csc \left(-\frac{\pi}{3}\right)-\frac{10 \sqrt{3}}{3}-\frac{10 \pi}{9}-2$$

Using the fact that $\csc \left(-\frac{\pi}{3}\right)=-\sqrt{3}$, we can simplify the expression.

$$f\left(-\frac{\pi}{3}\right)=-5 \cdot -\sqrt{3}-\frac{10 \sqrt{3}}{3}-\frac{10 \pi}{9}-2$$

$$f\left(-\frac{\pi}{3}\right)=5 \sqrt{3}-\frac{10 \sqrt{3}}{3}-\frac{10 \pi}{9}-2$$

$$f\left(-\frac{\pi}{3}\right)=\frac{15 \sqrt{3}}{3}-\frac{10 \sqrt{3}}{3}-\frac{10 \pi}{9}-2$$

$$f\left(-\frac{\pi}{3}\right)=\frac{5 \sqrt{3}}{3}-\frac{10 \pi}{9}-2$$

Now that we have the value of $f\left(-\frac{\pi}{3}\right)$, we can substitute it into the equation of the tangent line.

$$y-\left(\frac{5 \sqrt{3}}{3}-\frac{10 \pi}{9}-2\right)=5\left(x-\left(-\frac{\pi}{3}\right)\right)$$

Simplifying the equation, we get:

$$y-\frac{5 \sqrt{3}}{3}+\frac{10 \pi}{9}+2=5x+\frac{5 \pi}{3}$$

$$y=5x+\frac{5 \pi}{3}+\frac{5 \sqrt{3}}{3}-\frac{10 \pi}{9}-2$$

$$y=5x+\frac{5 \pi}{3}+\frac{5 \sqrt{3}}{3}-\frac{10 \pi}{9}-\frac{18}{9}$$

$$y=5x+\frac{5 \pi}{3}+\frac{5 \sqrt{3}}{3}-\frac{10 \pi}{9}-\frac{2}{1}$$

$$y=5x+\frac{5 \pi}{3}+\frac{5 \sqrt{3}}{3}-\frac{10 \pi}{9}-\frac{18}{9}$$

$$y=5x+\frac{5 \pi}{3}+\frac{5 \sqrt{3}}{3}-\frac{28}{9}$$

Conclusion

In this article, we found the equation of the tangent line to the graph of $f(x)=-5 \csc (x)-\frac{10 \sqrt{3}}{3}-\frac{10 \pi}{9}-2$ at $x=-\frac{\pi}{3}$. We first found the derivative of the function, then evaluated it at the given point to find the slope of the tangent line. Finally, we used the point-slope form of a line to find the equation of the tangent line. The equation of the tangent line is given by:

$$y=5x+\frac{5 \pi}{3}+\frac{5 \sqrt{3}}{3}-\frac{28}{9}$$

This equation represents the tangent line to the graph of the given function at the point $x=-\frac{\pi}{3}$.

Q: What is the tangent line to a curve?

A: The tangent line to a curve at a point is a line that just touches the curve at that point and has the same slope as the curve at that point.

Q: How do I find the equation of the tangent line to a curve?

A: To find the equation of the tangent line to a curve, you need to follow these steps:

1. Find the derivative of the function.
2. Evaluate the derivative at the given point to find the slope of the tangent line.
3. Use the point-slope form of a line to find the equation of the tangent line.

Q: What is the point-slope form of a line?

A: The point-slope form of a line is given by:

$$y-y_1=m(x-x_1)$$

where $m$ is the slope of the line, and $(x_1,y_1)$ is a point on the line.

Q: How do I find the derivative of a function?

A: To find the derivative of a function, you need to use the rules of differentiation, such as the power rule, the product rule, and the quotient rule.

Q: What is the power rule of differentiation?

A: The power rule of differentiation states that if $f(x)=x^n$, then $f'(x)=nx^{n-1}$.

Q: What is the product rule of differentiation?

A: The product rule of differentiation states that if $f(x)=u(x)v(x)$, then $f'(x)=u'(x)v(x)+u(x)v'(x)$.

Q: What is the quotient rule of differentiation?

A: The quotient rule of differentiation states that if $f(x)=\frac{u(x)}{v(x)}$, then $f'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{v(x)^2}$.

Q: How do I evaluate the derivative at a given point?

A: To evaluate the derivative at a given point, you need to substitute the value of $x$ into the derivative and simplify the expression.

Q: What is the significance of the tangent line to a curve?

A: The tangent line to a curve is significant because it represents the instantaneous rate of change of the function at a given point. It is also used to approximate the value of the function at a nearby point.

Q: Can I use the tangent line to a curve to make predictions about the behavior of the function?

A: Yes, you can use the tangent line to a curve to make predictions about the behavior of the function. However, you need to be careful not to extrapolate the behavior of the function beyond the point where the tangent line is valid.

Q: How do I determine the validity of the tangent line to a curve?

A: To determine the validity of the tangent line to a curve, you need to check if the function is differentiable at the given point. If the function is not differentiable at the point, then the tangent line is not valid.

Q: What is the relationship between the tangent line to a curve and the curve itself?

A: The tangent line to a curve is a line that just touches the curve at a given point and has the same slope as the curve at that point. The curve and the tangent line are related in that the tangent line represents the instantaneous rate of change of the curve at the given point.

Q: Can I use the tangent line to a curve to find the equation of the curve?

A: No, you cannot use the tangent line to a curve to find the equation of the curve. The tangent line is a local approximation of the curve, and it does not provide information about the global behavior of the curve.

Q: How do I use the tangent line to a curve to make approximations about the curve?

A: To use the tangent line to a curve to make approximations about the curve, you need to use the point-slope form of a line and the fact that the tangent line represents the instantaneous rate of change of the curve at a given point.