Respuesta :
Answer:
a) The increasing intervals would be from -4 to 0 and 4 to infinity. The decreasing interval would just be from negative infinity to -4 and 0 to 4.
b) The local maximum comes at x = 0. The local minimums would be x = -4 and x = 4
c) The inflection points are x= +/-√16/3
Step-by-step explanation:
To find the intervals of increasing and decreasing, we can start by finding the answers to part b, which is to find the local maximums and minimums. We do this by taking the derivatives of the equation.
f(x) = x^4 - 32x^2 + 2
f'(x) = 4x^3 - 64x
Now we take the derivative and solve for zero to find the local max and mins.
f'(x) = 4x^3 - 64x
0 = 4x^3 - 64x
0 = 4x(x + 4)(x - 4)
x = -4 OR x = 4 OR 0
Given the shape of a positive 4th power function function, we know that the first and last would be a minimums and the second would be a maximum.
As for the increasing, we know that a 4th power, positive function starts up and decreases to the local minimum. It also decreases after the local max. The rest of the time it would be increasing.
In order to find the inflection point, we take a derivative of the derivative and then solve for zero.
f'(x) = 4x^3 - 64x
f''(x) = 12x^2 - 64
0 = 12x^2 - 64
64 = 12x^2
16/3 = x^2
+/- √16/3 = x
-3/2 = x
Answer:
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:
[tex]f(x)=x^{4}-32x^{2} +3\\f'(x)= 4x^{4-1} -32(2)x^{2-1} + 0\\f'(x) = 4x^{3} -64x[/tex]
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:
[tex]f'(x) = 4x^{3} -64x=0\\x(4x^{2} -64)=0\\[/tex]
Applying the null factor property, we can equal to zero both factors:
[tex]x = 0\\4x^{2} -64=0\\4x^{2} =64\\x^{2} = \frac{64}{4} =16\\x = ± 4[/tex].
So, there are three critic points: -4; 0 ; 4, which give us four intervals.
[tex](- \infty; -4], (-4;0],(0;4],(4;+\infty)\\[/tex].
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:
[tex]f'(-5) = 4(-5)^{3} -64(-5) = -180 <0\\f'(-1) = 4(-1)^{3} -64(-1) = 60>0\\f'(5) = 180>0\\f'(1) = -60<0[/tex]
Therefore, the increasing intervals of the function are [tex](-4;0],(4;+\infty)\\[/tex]. On the other hand, the decreasing intervals are [tex](- \infty; -4],(0;4],\\[/tex].
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:
[tex]f"(x) = 12x^{2} -64=0\\12x^{2} =64\\x^{2} =\frac{64}{12} \\x=\sqrt{\frac{64}{12} } =±2.3[/tex]
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: [tex](- \infty;-2.3);(-2.3;2.3);(2.3:+\infty)[/tex]
We can use -3, 0 and 3 to evaluate each interval:
Replacing this values in the second derivative expression ([tex]f"(x) = 12x^{2} -64[/tex]), we have:
[tex]f"(-3) = 12(-3)^{2} -64=44>0\\f"(0)=-64<0\\f"(3) = 12(3)^{2} -64=44>0[/tex]
So, positive results mean concave up, negative results mean concave down. Therefore, [tex](- \infty;-2.3);(2.3:+\infty)[/tex] are concave up, and (-2.3;2.3) is concave down.
