y= (x^2)/ √(x - 4)   Domain. restricting your attention only to values of x for which f(x) is defined. Intercepts. Find intersections of the graph with the axes. The Y-intercept is (0, f(0)). To...

  1. y= (x^2)/ √(x - 4)

 

  1. Domain. restricting your attention only to values of x for which f(x) is defined.
  2. Intercepts. Find intersections of the graph with the axes. The Y-intercept is (0, f(0)). To obtain the X-intercept (c, 0) is harder and sometimes impossible. You need to find c such that f(c)= 0
  3. Symmetry. If f(x)=f(-x) , for example cos x = cos (-x), then the graph is symmetric with respect to the Y-axis. If f(x)= -f(-x), for example sin x= - sin (-x), then the graph is symmetric with respect to the origin.
  4. Asymptotes. Look for infinite limits and limits at infinity.
  5. Intervals of increase or decrease and local maxima and minima Reminder: Solve f’(x)> 0 and f’(x)<0 for monotonicity.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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1. You should notice that the radicand needs to be larger or equal to zero but since `sqrt(x-4)`  is also denominator of the fraction `(x^2)/(sqrt(x-4)),`  it cannot be equal to zero, such that:

`x - 4 gt 0 =gt x gt 4`

Hence, the domain of the function is represented by the interval `(4,oo).`

2. You need to find x and y intercepts such that:

`x = 0 =gt f(0) = (0^2)*sqrt(0-4) =gt f(0) = 0`

Hence, the graph of the function passes through origin (0,0).

`y = 0 =gt f(x) = 0`

You need to solve the equation `(x^2)/(sqrt(x-4)) = 0` .

Since `(sqrt(x-4))!=0` , then `x^2 = 0 =gt x = 0.`

Again, the fact that the graph passes through origin (0,0) is confirmed.

3. You need to test if the function is odd or even such that:

`f(-x) = (-x)^2sqrt(-x-4) =gt f(-x) = x^2sqrt(-x-4)`

Hence, the function is not odd, nor even, hence, the graph of the function is not symmetric with respect to origin or y axis.

4. You need to search for vertical asymptotes such that:

`lim_(x-gt4) (x^2)/(sqrt(x-4)) = 16/0 = +oo`

Hence, the graph of the function has vertical asymptote when x tends to 4.

You should search the horizontal asymptotes such that:

`lim_(x-gtoo) (x^2)/(sqrt(x-4)) = oo/oo`

You should use l'Hospital's theorem such that:

`lim_(x-gtoo) (4x)(sqrt(x-4)) = oo`

Hence, the function has no horizontal asymptotes.

You should search the oblique asymptotes  `y = mx+n`  such that:

`m = lim_(x-gtoo)(x^2)/(xsqrt(x-4))`

`m = lim_(x-gtoo) (x)/(sqrt(x-4)) = oo/oo`

m`= lim_(x-gtoo) 2(sqrt(x-4)) = oo`

Since m=oo, hence, the function has no oblique asymptotes.

5. You need to find derivative of the function using the quotient rule such that:

`f'(x) = (2xsqrt(x-4) - x^2/(2sqrt(x-4)))/(x-4)`

`f'(x) = (4x(x-4) - x^2)/(2(x-4)sqrt(x-4))`

`f'(x) = (4x^2 - 16x - x^2)/(2(x-4)sqrt(x-4))`

You need to solve the equation `f'(x) = 0`  to find the minimum and maximum points of the function such that:

`f'(x) = 0 =gt 3x^2 - 16x = 0 `

`x(3x - 16) = 0 =gt x = 0`

`3x -16 = 0 =gt 3x = 16 =gt x = 16/3`

Notice that f'(x) is negative if `x in (0;16/3)`  and it is positive if `x in (-oo;0)`  or `x in (16/3 ; oo).`

Hence, the function reaches its maximum at `x = 0`  and it reaches its minimum at `x = 16/3` .

The function increases over `(-oo,0) ` and `(16/3,oo)`  and it decreases over `(0,16/3).`

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