Find an equation of aline that is tangent to the curve f(x)+2cos3x and whose slope is a maximum?? Pleasee help!!
Oh, you're looking for a tangent line with maximum slope? That's like finding the trendiest line of all. Well, to find that line, we need to find the derivative of the curve first. So, prepare yourself for some mathematical magic!
Let's start by finding the derivative of f(x) + 2cos(3x). Taking the derivative of f(x) is easy peasy lemon squeezy, so we're just left with the cosine part. The derivative of cos(3x) is -3sin(3x) (chain rule, my friend).
Now, to find the tangent line with maximum slope, we need to find where the derivative is at its maximum or minimum. In this case, we're looking for a maximum. So, we set -3sin(3x) equal to zero and solve for x.
-3sin(3x) = 0
Since sin(3x) is equal to zero at x = 0, π, 2π, etc., we find that the possible x-values are 0, π/3, 2π/3, etc.
Now that we have the x-values, we can find the corresponding y-values by plugging them back into f(x) + 2cos(3x). And voila! You have pairs of x and y values.
Now, pick any of those pairs and use the point-slope equation, y - y1 = m(x - x1), to find the equation of the tangent line. The key here is to use the maximum slope you calculated earlier as the m value.
And there you have it! An equation of a line that is tangent to the curve f(x) + 2cos(3x) and whose slope is a maximum. It's like finding the hippest line in all of Mathland!
To find the equation of a line that is tangent to the curve f(x) + 2cos(3x) and has a maximum slope, we need to find the derivative of the given curve and then find the value of x that maximizes the derivative.
Step 1: Find the derivative of the given curve f(x) + 2cos(3x).
The derivative of f(x) is given by:
f'(x) = d/dx (f(x))
The derivative of 2cos(3x) with respect to x can be found using the chain rule. The chain rule states that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x).
In this case, g(x) = 3x and f(u) = 2cos(u). So, applying the chain rule, we get:
d/dx (2cos(3x)) = -2sin(3x) * 3 = -6sin(3x)
Therefore, the derivative of the given curve is:
f'(x) = -6sin(3x)
Step 2: Find the value of x that maximizes the derivative (-6sin(3x)).
To find the value of x that maximizes the derivative, we need to set the derivative equal to zero and solve for x.
-6sin(3x) = 0
Setting sin(3x) = 0, we get:
3x = nπ, where n is an integer.
Dividing by 3, we find:
x = nπ/3
So, the critical points (where the derivative is either maximum or minimum) are given by x = nπ/3.
Step 3: Find the maximum slope.
To find the maximum slope, we need to find the maximum value of the derivative at the critical points. Since the slope is the derivative, we need to find the maximum value of -6sin(3x).
Since sin(3x) has a maximum value of 1, the maximum derivative is:
f'(x) = -6sin(3x) = -6
Therefore, the maximum slope is -6.
Step 4: Find the equation of the tangent line.
Now that we have the slope of the tangent line (-6), we can find the equation of the tangent line using the point-slope form.
The point-slope form of a line is given by:
y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.
To find the equation of the tangent line, we need to know the point where the tangent line touches the curve.
Let's say the tangent line touches the curve at the point (a, f(a)). Therefore, the equation of the tangent line is:
y - f(a) = -6(x - a)
where f(a) is the value of the curve function at the point (a, f(a)).
So, the equation of the tangent line that is tangent to the curve f(x) + 2cos(3x) and has a maximum slope of -6 is:
y - f(a) = -6(x - a)
To find an equation of a line that is tangent to a curve and has a maximum slope, we need to follow these steps:
Step 1: Find the derivative of the given curve.
Step 2: Set the derivative equal to zero to find the critical points.
Step 3: Calculate the second derivative to determine the concavity of the curve at the critical points.
Step 4: Identify the critical point with the maximum slope.
Step 5: Use the point-slope form to find the equation of the tangent line.
Let's go through each step in detail:
Step 1: Find the derivative of the curve f(x) + 2cos(3x).
The derivative of f(x) with respect to x, denoted as f'(x), represents the slope of the curve at any given point. The derivative of cos(3x) with respect to x is -3sin(3x) since the derivative of cosine is negative sine and the chain rule is applied.
So, the derivative of f(x) + 2cos(3x) is f'(x) - 6sin(3x).
Step 2: Set the derivative equal to zero to find the critical points.
To find the critical points, we need to solve the equation f'(x) - 6sin(3x) = 0.
Step 3: Calculate the second derivative.
The second derivative, denoted as f''(x), indicates the concavity of the curve at a given point. To determine the second derivative, we differentiate the first derivative:
f''(x) = (f'(x) - 6sin(3x))' = f''(x) - 18cos(3x).
Step 4: Identify the critical point with the maximum slope.
Evaluate each critical point found in step 2 by plugging them into the second derivative, f''(x).
If f''(x) > 0, the curve is concave up at the point x.
If f''(x) < 0, the curve is concave down at the point x.
The critical point(s) that correspond to f''(x) > 0 will have a maximum slope.
Step 5: Use the point-slope form to find the equation of the tangent line.
Once the critical point with the maximum slope is determined, we can use the point-slope form of a line to find the equation of the tangent line. The point-slope form is given by:
y - y1 = m(x - x1),
where (x1, y1) denotes the coordinates of the critical point and m represents the slope.
That's it! By following these steps, you can find the equation of the line tangent to the given curve with the maximum slope.
I think f(x)=2cos3x.
f'(x)=-6sin3x
maxf'(x)=6 if sin3x=-1 when x=-Pi/6
f(-Pi/6)=0
The equation of tangent
y=6x+Pi