Mathematics, 01.10.2019 09:30 luxmimspcbcxf

Consider the equation below. (if an answer does not exist, enter dne.) f(x) = x4 − 32x2 + 3 (a) find the interval on which f is increasing. (enter your answer using interval notation.) find the interval on which f is decreasing. (enter your answer using interval notation.) (b) find the local minimum and maximum values of f. local minimum value local maximum value (c) find the inflection points. (x, y) = (smaller x-value) (x, y) = (larger x-value) find the interval on which f is concave up. (enter your answer using interval notation.) find the interval on which f is concave down.(enter your answer using interval notation.)

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Answer from: Dmoney7784

This is a problem where we need to analyse the function. To do so, we can recur to the applications of derivatives.

So, we have to derive the function first:

$f(x)=x^{4}-32x^{2} +3\\f'(x)= 4x^{4-1} -32(2)x^{2-1} + 0\\f'(x) = 4x^{3} -64x$

To analyse the function, we need to determinate the intervals using critic points, which can be found making the function equal to zero and then factorize:

$f'(x) = 4x^{3} -64x=0\\x(4x^{2} -64)=0\\$

Applying the null factor property, we can equal to zero both factors:

$x = 0\\4x^{2} -64=0\\4x^{2} =64\\x^{2} = \frac{64}{4} =16\\x = ± 4$ .

So, there are three critic points: -4; 0 ; 4, which give us four intervals.

$(- \infty; -4], (-4;0],(0;4],(4;+\infty)\\$ .

Now, we have to evaluate each intervals, to know if they are increasing or decreasing. If the result of each evaluation results more than zero, then it's increasing, if results less than zero, it's decreasing.

So, the test values are -5 (1st interval), -1 (2nd interval), 1 (3rd interval), 5 (4th interval). As you can see, each value is included in one interval. The evaluation can be done just by replacing each value:

$f'(-5) = 4(-5)^{3} -64(-5) = -180 0\\f'(1) = -60$

Therefore, the increasing intervals of the function are $(-4;0],(4;+\infty)\\$ . On the other hand, the decreasing intervals are $(- \infty; -4],(0;4],\\$ .

On the other hand, if the function change from negative to positive in c, then the function has a minimum located in (c ; f(c)). So, in -4 and 4, the function change from negative to positive (from decreasing to increasing), so the are minimums located in (-4;-253) and (4;-253). However, if the function change from positive to negative in c, then the functions has a maximum locate in (c ; f(c)). In this case, 0 changes from positive to negative. So, the maximum is located in (0; 3).

At last, the inflection points can be find using the second derivative criteria. First, we derive again the function, to find the second derivative, and then equal to zero to find inflexion points:

$f"(x) = 12x^{2} -64=0\\12x^{2} =64\\x^{2} =\frac{64}{12} \\x=\sqrt{\frac{64}{12} } =±2.3$

Therefore, the inflexion points are located in -2.3 and +2.3. Next, we do the same process, we determine the intervals, then we evaluate each of them to find which interval is concave up and which is concave down.

Intervals: $(- \infty;-2.3);(-2.3;2.3);(2.3:+\infty)$

We can use -3, 0 and 3 to evaluate each interval:

Replacing this values in the second derivative expression ( $f"(x) = 12x^{2} -64$ ), we have:

$f"(-3) = 12(-3)^{2} -64=440\\f"(0)=-64$

So, positive results mean concave up, negative results mean concave down. Therefore, $(- \infty;-2.3);(2.3:+\infty)$ are concave up, and (-2.3;2.3) is concave down.

Answer from: Quest

step-by-step explanation:

through just a quick inspection, one can see that the graph of this cubing function starts in quadrant iii and ends in quadrant i. when x = 0, y = 3, so the y-intercept is (0, 3).

if graphs are part of the problem statement, share them. .

Answer from: Quest

a² + b²= c²

24² + 7² = c²

c= 25

What is the value of x? enter your answer in the box. x =

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Consider the equation below. (if an answer does not exist, enter dne.) f(x) = x4 − 32x2 + 3 (a) find...