# Let f(x) = -x^{4} - 5x^{3} + 6x + 7. Find the open intervals on which f is concave up (down)....

## Question:

Let {eq}f(x) = -x^{4} - 5x^{3} + 6x + 7 {/eq}.

Find the open intervals on which {eq}f {/eq} is concave up (down).

Then determine the {eq}x {/eq}-coordinates of all inflection points of {eq}f {/eq}.

## Inflection Points and Concavity

Inflection points are the points where the concavity of the curve changes. It is computed by getting the values that make the second derivative zero or undefined.

Let's begin by getting all the possible inflection points using second derivative.

{eq}\displaystyle f(x) = -x^{4} - 5x^{3} + 6x + 7 \\ \displaystyle f'(x) = -4x^{3} - 15x^{2} + 6 \\ \displaystyle f''(x) = -12x^{2} - 30x {/eq}

Since this is a polynomial function, it is defined over the entire domain. There is no value that will make the second derivative undefined. So we will only look at the values that make the second derivative zero.

{eq}\displaystyle 0 = -12x^{2} - 30x \\ \displaystyle 0 = (-6x)(2x + 5) {/eq}

Apply zero product rule.

{eq}\displaystyle 0 = (-6x), 0 = (2x + 5) \\ \displaystyle \frac{0}{-6} = x, -5 = 2x \\ \displaystyle x = 0, \frac{-5}{2} = x \\ \displaystyle x = 0, x= -\frac{5}{2} {/eq}

These values are not necessarily inflection points. We need to check the intervals on the left and right side of the points and check if the concavity changes. Since there are two values, we will be looking at three intervals, which are {eq}\displaystyle \left (-\infty, -\frac{5}{2} \right ), \left (-\frac{5}{2}, 0 \right ) {/eq} and {eq}(0, +\infty) {/eq}. We will choose a point within these intervals and substitute it to the second derivative. If it results to a positive number then the interval is concave up. If the result is a negative number then the interval is concave down.

For the first interval {eq}\displaystyle \left (-\infty, -\frac{5}{2} \right ) {/eq}, ket's choose {eq}x = -3 {/eq}:

{eq}\displaystyle f''(x) = -12x^{2} - 30x \\ \displaystyle f''(-3) = -12(-3)^2 - 30(-3) \\ \displaystyle f''(-3) = -12(9) + 90 \\ \displaystyle f''(-3) = -108 + 90 \\ \displaystyle f''(-3) = -18 < 0 \therefore \left (-\infty, -\frac{5}{2} \right ) \text{ is concave down} \\ {/eq}

For the second interval {eq}\displaystyle \left (-\frac{5}{2},0 \right ) {/eq}, ket's choose {eq}x = -1 {/eq}:

{eq}\displaystyle f''(x) = -12x^{2} - 30x \\ \displaystyle f''(-1) = -12(-1)^2 - 30(-1) \\ \displaystyle f''(-1) = -12(1) + 30 \\ \displaystyle f''(-1) = -12 + 30 \\ \displaystyle f''(-1) = 18 > 0 \therefore \left ( -\frac{5}{2},0 \right ) \text{ is concave up} {/eq}

For the third and last interval {eq}\displaystyle \left (0, +\infty \right ) {/eq}, ket's choose {eq}x = 1 {/eq}:

{eq}\displaystyle f''(x) = -12x^{2} - 30x \\ \displaystyle f''(1) = -12(1)^2 - 30(1) \\ \displaystyle f''(1) = -12(1) - 30 \\ \displaystyle f''(1) = -12 - 30 \\ \displaystyle f''(1) = -32 < 0 \therefore \left ( 0, +\infty \right ) \text{ is concave down} {/eq}

In summary, the following open intervals are:

{eq}\displaystyle \text{concave up}: \left ( -\frac{5}{2},0 \right ) \\ \displaystyle \text{concave down}: \left (-\infty, -\frac{5}{2} \right ) \cup \left ( 0, +\infty \right ) {/eq}

The concavity changes at the possible inflection points. Hence, the x- coordinates of the inflection points are

{eq}\displaystyle x = -\frac{5}{2}, x = 0 {/eq} 