# Consider the function f(x) = 4 x^3 - 3 x^4. Find all I. Critical points of f(x). II. The...

## Question:

Consider the function {eq}\displaystyle (x) = 4 x^3 - 3 x^4. {/eq} Find all
I. Critical points of {eq}f(x). {/eq}
II. The intervals on which {eq}f(x) {/eq} is concave up / down.
III. Points of inflection of {eq}f(x).{/eq}
IV. Analyse critical points to determine which are local minimums and which are local maximums.

## Analyzing Functions with Derivatives

If we can find the first or second derivative of a function, we can calculate a lot of information about this function. This includes where a function has a local maximum or minimum, since the first derivative can tel us where a function is increasing or decreasing, and also where a function has inflection points, since the second derivative can inform us about concavity.

In order to complete the parts of this problem, we need the first and second derivative of this function. These can be found quickly using the power rule, so let's compute these now.

{eq}f'(x) = 12x^2 - 12x^3\\ f''(x) = 24x - 36x^2 {/eq}

I. The critical points occur when the first derivative has a value of zero, so we can find these points by setting the derivative equal to zero.

{eq}12x^2 - 12x^3 = 0\\ 12x^2(1-x) = 0\\ x = 0\\ x = 1 {/eq}

II. The function is concave up when the second derivative is positive and concave down when the second derivative is negative. Let's see where the concavity could switch by setting the second derivative equal to zero.

{eq}24x - 36x^2 = 0\\ 12x(2-3x) = 0\\ x = 0\\ x = \frac{2}{3} {/eq}

We can now see if the function is concave up or concave down outside of these points by testing the second derivative somewhere in the intervals.

{eq}f''(-1) = 24(-1) - 36(-1)^2 = -60\\ f''(0.5) = 24(0.5) - 36(0.5)^2 = 3\\ f''(1) = 24(1) - 36(1)^2 = -12 {/eq}

Thus, the function is concave up on the interval {eq}(0, \frac{2}{3}) {/eq} and concave down on the intervals {eq}(-\infty, 0) {/eq} and {eq}(\frac{2}{3}, \infty) {/eq}.

III. Since the function changes concavity at both of the places where the second derivative is zero, these are the inflection points of the function: {eq}x = 0 {/eq} and {eq}x = \frac{2}{3} {/eq}.

IV. In order to find where this function has local maximums and local minimums, we can analyze the critical points we found. We have the second derivative already, so let's use the second derivative test.

{eq}f''(0) = 24(0) - 36(0)^2 = 0\\ f''(1) = 24(1) - 36(1)^2 = -12 {/eq}

The second derivative is zero at our first critical point, so that is neither a local maximum nor a local minimum. However, it's negative at our second critical point, so the function has a local maximum there. This means that {eq}x = 1 {/eq} is a local maximum of the function.