# How do you find the intervals of increasing and decreasing using the first derivative given #y=10(5-sqrt(x^2-3x+16))#?

Increasing

Decreasing

As can be seen from the diagram, maximum/minimum and points of inflection( saddle points) have a gradient of zero. These are sometimes referred to as stationary points ( points where the function is neither increasing nor decreasing ).The gradient on either side of these points can be used to determine where the function is increasing or decreasing.

We first find the derivative of the function:

We now solve this for zero:

So stationary point at

We now test either side of this point. We can see from the diagram, that when the function is increasing it has a positive gradient and when decreasing it has a negative gradient. We need to take values near the stationary point and then plug them into the derivative.

To the left:

Let

To the right:

Let

Increasing

Decreasing

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The graph is increasing until

So your goal is to find the intervals of increasing and decreasing, which essentially means you're trying to find where the instantaneous slopes are increasing or decreasing, which is the definition of a derivative: Giving you the instantaneous rate of change at any given point. You're essentially looking for:

The first rule I'm going to use is the "Derivative of a constant times a function," which states

Therefore, I can get

Next, we need to take the derivative. The 5 is immediately removed, because the derivative of a constant is 0, and therefore we are left with:

Again, we can take out the negative, because that's still the constant times a function:

Now, we can use the chain rule, which says:

In this case, the following statements are true:

Therefore, we can plug in everything into the chain rule to get:

Notice: I multiplied the -10 in just to clean up the question a bit.

Next, we want to solve for 0, which will tell us when the function isn't increasing or decreasing. Thankfully, because this is a rational function, all we need to do is solve for the numerator and make sure the denominator isn't 0.

So we know for values less than 1.5, the function is increasing. Now, we'll take one on the other side. I'll take x=3.

Therefore, the function is increasing before 1.5, and decreasing after, and isn't doing either at 1.5.

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To find the intervals of increasing and decreasing using the first derivative, follow these steps:

- Find the first derivative of the function ( y = 10(5 - \sqrt{x^2 - 3x + 16}) ).

( y' = \frac{dy}{dx} = -\frac{10x - 15}{\sqrt{x^2 - 3x + 16}} ).

- Set the first derivative equal to zero and solve for ( x ) to find the critical points.

( -\frac{10x - 15}{\sqrt{x^2 - 3x + 16}} = 0 ).

( 10x - 15 = 0 ).

( x = \frac{15}{10} = \frac{3}{2} ).

- Determine the intervals between critical points and plug test points into the first derivative to determine the sign (positive or negative) in each interval.

Test ( x = 0 ): ( y' = -\frac{10(0) - 15}{\sqrt{16}} = -\frac{15}{4} < 0 ). Test ( x = 3 ): ( y' = -\frac{10(3) - 15}{\sqrt{3^2 - 3(3) + 16}} = \frac{15}{\sqrt{16}} > 0 ).

- Thus, the function is decreasing on the interval ( (-\infty, \frac{3}{2}) ) and increasing on the interval ( (\frac{3}{2}, \infty) ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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