Question

How do you sketch the graph `y=x^5-16x` using the first and second derivatives?

Answer 1

graph{x^5 - 16x [-40, 40, -20, 20]}

Explanation:

The first derivative tells us where the function is either decreasing or increasing (increasing where the first derivative > 0 and decreasing where the first derivative < 0. Note that when first derivative = 0, this gives the x-coordinate of the turning point). Moreover, this test can help us locate the critical/stationary points of the function.

The second derivative test tells us more about the concavity of the function and can help us find the points of inflection of the function. Where the 2nd derivative > 0: the function is concave up. Where the 2nd derivative < 0: the function is concave down.

The first derivative of y is 5x^4 - 16. We want to know where the function is increasing , so we'll set up an inequality from the first derivative, that is, `5x^4 - 16 > 0`. Solving this inequality, the result is `x > 2/(5^(1/4))` or `x < 2/(5^(1/4))`. This means that the function is increasing where `x > 2/(5^(1/4))`.

For the next part, we consider where the first derivative is less than 0. After computing this inequality, you will observe that the function is decreasing where `-2/(5^(1/4))< x < 2/5^(1/4)`.

To find the turning points: set the first derivative equal to 0 and solve. The result is `x = 2/5^(1/4)` or `x =-2/5^(1/4)` . To find the corresponding y-values substitute the x-values we just derived in the ORIGINAL equation (Do not substitute both values at the same time! Choose one value and plug it in wherever you have x to get y, then plug in the next x-value to get the next y-value). Thus you have your turning points!

We now know where the function is increasing, decreasing, and we have its turning points.

The second derivative of y is `20x^3`. Compute the solution where the 2nd derivative > 0 and the answer is simply x > 0. Similarly, the 2nd derivative is < 0 when x < 0. Now we take all this information and graph it.