# Calculating Curve Tangent at x=-π/4

• SwedishFred
In summary, the problem is to find the equation of the line that is tangent to the curve at x = -π/4. The given equation is f(x) = 1/3 sin(3x-π/4). The attempted solution uses the tangent equation and calculates the slope and y-intercept of the tangent line, but there is an error in the calculation due to using the wrong mode on a calculator or using an approximation for π. The correct answer is not 3.33*10^-14, and the question should be posted in the Calculus & Beyond section.
SwedishFred

## Homework Statement

Deside curve tangent in point x=-π/4

## Homework Equations

f(x)=1/3sin(3x-π/4)
y=f(x)

## The Attempt at a Solution

f(-π/4)=-1
using the tangent equation
y=kx+m
y=-1*(-π/4)+m
y=1/3sin(3(-π/4)-π/4)
≈3.33*10^-14
3.33*10^-14=-1*(-π/4)+m
f(x)≈-1*(-π/4)+0,79
is this right ? I have a bad feeling...

SwedishFred said:

## Homework Statement

Deside curve tangent in point x=-π/4
Deside?
What is that?
Is the problem to find the line that is tangent to the curve at x = -π/4?
SwedishFred said:

## Homework Equations

f(x)=1/3sin(3x-π/4)
y=f(x)

## The Attempt at a Solution

f(-π/4)=-1
using the tangent equation
y=kx+m
y=-1*(-π/4)+m
y=1/3sin(3(-π/4)-π/4)
≈3.33*10^-14
No.
When x = -π/4, y = 1/3 * sin(3(-π/4)-π/4)) = 1/3 * sin(-π). This is one of the angles whose sine and cosine you should have committed to memory. It looks like you might be using a calculator to do this, and either have the calculator in the wrong mode (it should be in radian mode) or you are using an approximation to π. Either way will not give you the right answer.

SwedishFred said:
3.33*10^-14=-1*(-π/4)+m
f(x)≈-1*(-π/4)+0,79
is this right ? I have a bad feeling...

By the way, questions that involve taking derivatives should be posted in the Calculus & Beyond section, not in the Precalculus section.

## 1. How do you calculate the curve tangent at x=-π/4?

To calculate the curve tangent at x=-π/4, you will need to use the first derivative of the function at that point. This can be found by taking the limit as h approaches 0 of [f(-π/4+h)-f(-π/4)]/h. Once you have the derivative, plug in x=-π/4 to get the slope of the curve at that point.

## 2. What is the significance of calculating the curve tangent at x=-π/4?

Calculating the curve tangent at x=-π/4 allows you to determine the slope or rate of change of the function at that specific point. This can be useful in analyzing the behavior of the function and making predictions about its future values.

## 3. Can you explain the concept of a tangent line in the context of calculating at x=-π/4?

A tangent line is a line that touches a curve at a single point, in this case x=-π/4. It represents the instantaneous rate of change of the function at that point. By calculating the curve tangent at x=-π/4, you are finding the slope of this tangent line.

## 4. What if the function is not differentiable at x=-π/4? Can the curve tangent still be calculated?

If the function is not differentiable at x=-π/4, then the curve tangent cannot be calculated as the first derivative does not exist at that point. This could be due to a sharp turn or a cusp in the curve at that point.

## 5. How can knowing the curve tangent at x=-π/4 be applied in real-world situations?

Knowing the curve tangent at x=-π/4 can be applied in various fields such as physics, engineering, and economics. For example, in physics, it can be used to determine the velocity of a moving object at a specific point in time. In economics, it can be used to analyze the growth rate of a company's profits. Essentially, it can be used to understand the instantaneous rate of change of a phenomenon.

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