contestada

1) Let f(x)=6x+6/x. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relative maxima (minima).



Let f(x)=6−4/x+2/x^2. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relative maxima (minima).



Let f(x)=(8x^2)/(x−4). Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relative maxima (minima).



Let f(x)=6(x−2)^(2/3) +4. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relative maxima (minima).



Let f(x)=8√x −6x for x>0. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relative maxima (minima).

Respuesta :

erturkmemmedli

Answer:

1) increasing on (-∞,-1] ∪ [1,∞), decreasing on [-1,0) ∪ (0,1]

[tex]x = -1[/tex] is local maximum, [tex]x = 1[/tex] is local minimum

2) increasing on [1,∞), decreasing on (-∞,0) ∪ (0,1]

[tex]x = 1[/tex] is absolute minimum

3) increasing on (-∞,0] ∪ [8,∞), decreasing on [0,4) ∪ (4,8]

[tex]x = 0[/tex] is local maximum, [tex]x = 8[/tex] is local minimum

4) increasing on [2,∞), decreasing on (-∞,2]

[tex]x = 2[/tex] is absolute minimum

5) increasing on the interval (0,4/9], decreasing on the interval [4/9,∞)

[tex]x = 0[/tex] is local minimum, [tex]x = 4/9[/tex] is absolute maximum

Step-by-step explanation:

To find minima and maxima the of the function, we must take the derivative and equalize it to zero to find the roots.

1) [tex]f(x) = 6x + 6/x[/tex]

[tex]f\prime(x) = 6 - 6/x^2 = 0[/tex] and [tex]x \neq 0[/tex]

So, the roots are [tex]x = -1[/tex] and [tex]x = 1[/tex]

The function is increasing on the interval (-∞,-1] ∪ [1,∞)

The function is decreasing on the interval [-1,0) ∪ (0,1]

[tex]x = -1[/tex] is local maximum, [tex]x = 1[/tex] is local minimum.

2) [tex]f(x)=6-4/x+2/x^2[/tex]

[tex]f\prime(x)=4/x^2-4/x^3=0[/tex] and [tex]x \neq 0[/tex]

So the root is [tex]x = 1[/tex]

The function is increasing on the interval [1,∞)

The function is decreasing on the interval (-∞,0) ∪ (0,1]

[tex]x = 1[/tex] is absolute minimum.

3) [tex]f(x) = 8x^2/(x-4)[/tex]

[tex]f\prime(x) = (8x^2-64x)/(x-4)^2=0[/tex] and [tex]x \neq 4[/tex]

So the roots are [tex]x = 0[/tex] and [tex]x = 8[/tex]

The function is increasing on the interval (-∞,0] ∪ [8,∞)

The function is decreasing on the interval [0,4) ∪ (4,8]

[tex]x = 0[/tex] is local maximum, [tex]x = 8[/tex] is local minimum.

4) [tex]f(x)=6(x-2)^{2/3} +4=0[/tex]

[tex]f\prime(x) = 4/(x-2)^{1/3}[/tex] has no solution and [tex]x = 2[/tex] is crtitical point.

The function is increasing on the interval [2,∞)

The function is decreasing on the interval (-∞,2]

[tex]x = 2[/tex] is absolute minimum.

5) [tex]f(x)=8\sqrt x - 6x[/tex] for [tex]x>0[/tex]

[tex]f\prime(x) = (4/\sqrt x)-6 = 0[/tex]

So the root is [tex]x = 4/9[/tex]

The function is increasing on the interval (0,4/9]

The function is decreasing on the interval [4/9,∞)

[tex]x = 0[/tex] is local minimum, [tex]x = 4/9[/tex] is absolute maximum.

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