Finding critical points of functions and using the sign test.

Question

I was given a question with three parts where each part is a separate function that I need to find the critical points, inflection points, concavity and sketch the graph.

Part A) f(x)= x^4 +2x^3 -1

I know that I need to find the first derivative so I did and factored and I got

2x^2 (2x+3)

I'm not sure how to find critical points or inflection points. Would I try and solve for x?

then to find if the function is increasing or decreasing I know I need to use a sign test but the videos I have watched and my teachers example makes no sense. what exactly constitutes the numbers on the line? and how do you know if it is a plus or minus between them?

Then for concavity I know that the second derivative is used to find that so I took

12x^2 + 12x which I factored to get 12x(x+1)

My teacher also used the sign test for this, but that confuses me. She also calls this test the first derivative test so wouldn't it only be applicable to the first derivative?

If you could please explain the process to me I am sure I can figure out the other parts to this problem. Thank you!

 

Answer 1

Begin the function f(x) = x⁴ + 2x³ - 1

 

Take the first and second derivative of this function.

 

f'(X) = 4x³ + 6x² = 2x² *(2x + 3)

f"(x) = 12x² + 12x = 12x (x + 1)

 

The critical values are in this case the values of x for which f'(x) = 0.

Here 2x² * ( 2x + 3) = 0 and the critical values ar x = 0 or x = -1.5

At these values the slope of the tangent line through the point is zero.

 

The first derivative represents the slope of the tangent line at the given point.

For the value of  x = 0, take a small value of x less than zero and see what its slope is.

In this case for instance let this value be say -.01 and its slope is

f'(-.01) = 2 (-.01)² * (2*(-.01) + 3) = 0.000596.

This means that the slope of the tangent of line is positive for the values of x less than 0 and close to 0

When the slope is positive, then the tangent lines are increasing.

 

Also, take a small value of x greater than 0, for instance .01 and find

the slope of the line with this value.

f'(.01) = 2(.01)2*(2*.01+3) = 0.000604

This means that the slope of the tangent of line is positive for the values of x greater than and close to 0
When the slope is positive, then the tangent lines are increasing.

 

For the critical value of x = 0, the tangent lines less than zero and also the tangent lines greater than zero

are both positive.  This is an indication that around zero, the function is increasing and means that the value

could be an inflection point.

 

Do the same thing with the other critical value, x = -1.5

Take a value less than -1.5, say -1.51 and find the slope

f'(-1.51) = -0.091204, so the slope of the tangent lines less than and close to -1.5 are negative and the function is decreasing.

 

Take a value greater than -1.5, say -1.49 and find the slope

f'(-1.49) = 0.08804, so the slope of the tangent lines greater than and close to -1.5 are positive and the function is increasing.

 

This case can be represented by (-, +), where - indicates the slope of tangent lines less than a critical value is negative and slope of tangent lines greater than the critical value are positive.  When this happens, the critical value is a relative minimum.  This is true in this example for x = -1.5, it is a relative minimum.

 

When the slopes less than and greater than a critical value are represented by (+, -), then the critical value is a relative maximum.

 

The case of x = 0 represents the final case when the sign of the slope does not change when values less than or greater the critical value is taken.  If the slope is positive, then the function is increasing near the critical value, and when the slope is negative the function is decreasing near the critical value.

 

The possible points of inflection have the value of x when f"(0) = 0.

Here 12x * (x + 1) = 0 and the point of inflections occur when x = 0 and when x = -1.

 

If values less than or greater than an inflection point have positive second derivative values in that area than the function is concave up in that area.

If values less than or greater than an inflection point have negative second derivative values in that area than the function is concave down in that area.

 

Take the inflection point x = 0 in this example.

For a value less than 0, say x = -.01, f"(-.01) = 12 * (-.01) * (-.01 + 1) = -0.1188.

Then the function is concave down to the left (less than) zero.

 

For a value greater than 0, say x = .01, f"(.01) = 12 * (.01) * (.01 + 1) = 0.1212

Then the function is concave up to the right (greater than) zero.

 

Take the inflection point x = -1 in this example.

For a value to the left of -1, say x = -1.01, f"(-1.01) = 0.1212

Then the function is concave up to the left of -1.

 

For a value to the right of -1, say x = -.99, f"(-.99) = -0.1188

Then the function is concave down to the right of -1.