consider the following.
Function
y = cos 3x
point (pi/4, -square root of 2/2)
(a) Find an equation of the tangent line to the graph of f at the given point. (Let x be the independent variable and y be the dependent variable.)
y’= m (the slope) of the tangent line
y' = -3sin(3x)
so, at x=π/4, y = -3sin(3π/4) = -3(√2/2)
Then plug that into point slope formula: y-y1 = m (x-x1)
y-(-√2/2)= 3√2/2 (x- π/4)
I believe Steve’s work is right except that sin (3π/4)= √2/2, not 1/2
To find the equation of the tangent line to the graph of f at the given point, we need to determine the slope of the tangent line.
The slope of the tangent line can be found using the derivative of the function f(x).
Given function: y = cos(3x)
To find the derivative of the function, we differentiate with respect to x using the chain rule:
d/dx (y) = d/dx (cos(3x))
Using the chain rule, we have:
d/dx (y) = -sin(3x) * d/dx (3x)
Recall that d/dx (3x) = 3 * d/dx (x) = 3.
Therefore, the derivative of the function is:
dy/dx = -3sin(3x)
Now, we need to find the slope of the tangent line at the given point (pi/4, -√2/2).
Substitute x = pi/4 into dy/dx:
dy/dx = -3sin(3(pi/4)) = -3sin(3pi/4) = -3sin(3pi/4) = -3*(-√2/2) = 3√2/2
So, the slope of the tangent line at the point (pi/4, -√2/2) is 3√2/2.
Now we can use the point-slope form of a linear equation to find the equation of the tangent line.
y - y1 = m(x - x1)
Substituting the values of the point (pi/4, -√2/2) and the slope (3√2/2) into the equation, we have:
y - (-√2/2) = (3√2/2)(x - pi/4)
Simplifying:
y + √2/2 = (3√2/2)(x - pi/4)
Multiply both sides by 2 to eliminate the fractions:
2y + √2 = 3√2(x - pi/4)
Finally, rearrange to get the equation in the general form:
2y = 3√2(x - pi/4) - √2
2y = 3√2x - 3√2(pi/4) - √2
2y = 3√2x - (3√2pi + √2)/2
So, the equation of the tangent line to the graph of f at the point (pi/4, -√2/2) is 2y = 3√2x - (3√2pi + √2)/2.
To find the equation of the tangent line to the graph of the function f at the given point, we need to determine the slope of the tangent line and then use the point-slope form of a linear equation.
Step 1: Calculate the derivative of the function f to find the slope of the tangent line at any given point.
In this case, f(x) = cos(3x). To find the derivative, apply the chain rule: d/dx(cos(u)) = -sin(u) * du/dx.
Since u = 3x, du/dx = 3. Therefore, the derivative of f(x) is f'(x) = -3sin(3x).
Step 2: Plug in the x-coordinate of the given point into the derivative to find the slope.
The x-coordinate of the given point is pi/4, so the slope of the tangent line is f'(pi/4) = -3sin(3(pi/4)).
Step 3: Calculate the value of -3sin(3(pi/4)).
First, simplify the argument of the sine function: 3(pi/4) = 3pi/4. Then, take the sine of this value: sin(3pi/4) = -sqrt(2)/2.
Finally, multiply by the coefficient -3 to get -3sin(3(pi/4)) = -3 * (-sqrt(2)/2) = 3sqrt(2)/2.
Step 4: Use the point-slope form of a linear equation, y - y1 = m(x - x1), where m is the slope and (x1, y1) is the given point.
The slope, m, is 3sqrt(2)/2, and the point is (pi/4, -sqrt(2)/2).
Plugging these values into the point-slope form, we get:
y - (-sqrt(2)/2) = (3sqrt(2)/2)(x - pi/4).
Simplifying further, we can rewrite this equation as:
y + sqrt(2)/2 = (3sqrt(2)/2)(x - pi/4).
Hence, the equation of the tangent line to the graph of f at the given point is:
y = (3sqrt(2)/2)(x - pi/4) - sqrt(2)/2.
first, find the slope of the tangent line anywhere:
y' = -3sin(3x)
so, at x=π/4, y = -3sin(3π/4) = -3/√2
Now you have a point and a slope, so the line is
y + 1/√2 = -3/√2 (x - π/4)