Determining the Increasing or Decreasing Nature of y sin x^5 at x 15 Degrees

Is the Graph of y sin x5 Increasing or Decreasing at x 15 Degrees?

To determine whether the graph of y sin x5 is increasing or decreasing at x 15 degrees, we need to find the first derivative of the function and evaluate it at that point.

Step-by-Step Solution

1. Find the First Derivative

The function is y sin x5. Using the chain rule, we can find the derivative as follows:

(frac{dy}{dx}  cos x^5 cdot frac{d}{dx}x^5)

The derivative of x^5 is 5x^4. Therefore, the first derivative of the function is:

(frac{dy}{dx}  cos x^5 cdot 5x^4)

2. Evaluate the Derivative at x 15 Degrees

Evaluating the derivative at x 15 degrees, we get:

(frac{dy}{dx}Bigg|_{x15}  cos(15^5) cdot 5(15^4))

First, calculate 15^5:

(15^5  759375)

Next, evaluate cos(759375) using a calculator or numerical software. To simplify, we can find the equivalent angle modulo 2pi (or 6.2832):

(759375 mod 2pi approx 759375 mod 6.2832 approx 1.293)

Now evaluate cos(1.293) using a calculator:

(cos(1.293) approx 0.267)

Next, calculate 5(15^4):

(15^4  50625 rightarrow 5 cdot 50625  253125)

Combining these results, we get:

(frac{dy}{dx}Bigg|_{x15} approx 0.267 cdot 253125 approx 67601.875)

Since the derivative is positive, the graph of y sin x5 is increasing at x 15 degrees.

Additional Insights

The function y sin x5 has an exponentially increasing frequency with a constant wavelength. As x moves to the far right, the sine wave gets steeper and steeper, making the function appear as a solid bar from y 1 to y -1, extending to infinity. The waves are so close together that the graph rapidly oscillates between these bounds.

The critical points of the function can be found by solving:

(cos x^5  0)

This occurs when:

(x  left(frac{npi}{2}right)^{1/5})

However, for practical purposes, we can focus on the behavior around specific points, such as x 15 degrees, without solving for all critical points.

Conclusion

The graph of y sin x5 is increasing at x 15 degrees, as the first derivative evaluated at this point is positive.