For a certain function, f'(x)=3+x. For what value of x does the function have a stationary point? What type of stationary point is it?

I've found x to be -3, but how do I determine the type of stationary point?

You know that if f'(x) = 3+x , then

f(x) = 3x + (1/2)x^2+ c
which is a parabola opening upwards, so when x = -3 you have a vertex
and that vertex must be a minimum point.

f'(x)=3+x

you know that f'(-3) = 0
for x < -3, f' < 0
for x > -3, f' > 0
so, f(x) is falling, then stationary, then rising
Looks like a minimum to me.

To determine the type of stationary point, you need to analyze the second derivative of the function. The second derivative will provide information about the concavity of the function at the stationary point.

Given that f'(x) = 3 + x, we can find the second derivative, f''(x), by differentiating f'(x) with respect to x:

f''(x) = d/dx (3 + x) = 1

The second derivative is a constant, which means it doesn't depend on x.

Now, let's analyze the type of stationary point using the second derivative:

1. If f''(x) > 0, then the function is concave up at the stationary point, indicating a minimum.
2. If f''(x) < 0, then the function is concave down at the stationary point, suggesting a maximum.
3. If f''(x) = 0 or undefined, then the test is inconclusive.

In this case, f''(x) is equal to 1, which is greater than 0. Therefore, the function has a stationary point at x = -3, and it is a minimum.

So, the answer is x = -3, and the stationary point is a minimum point.