3. Consider the function f(x) = (2x^2 - 18)/(x^2

3. Consider the function \(f(x) = \frac{2x^2 - 18}{x^2 - x - 2}\) (a) Find out where \(f(x)\) crosses the x-axis and the y-axis (if it does). (b) Determine the vertical and horizontal asymptotes of \(f(x)\) (if any). (c) Differentiate \(f(x)\). (d) Determine the points where the derivative of f is zero (if any) and the intervals(s) of increase and decrease of \(f(x)\). (e) Sketch the graph of \(f(x)\) (no need to be very precise: your sketch only needs to reflect the properties of f determined in parts (a), (b), and (d)).

Transcript

00:01 Hi there, in this question we are given this function f of x is equal to e to the power x over 2 and divided by x.

00:08 So the first one, so we have to find is there in a y intercept? so for the y intercept, we have to plug in zero, so we need to just find f of zero, which is equal to e to the power 0 divided by 0, which is equal to 1 over 0, which is infinite.

00:27 It.

00:27 So that means there is no any y intercept for this function.

00:32 So by the way, x is equal to would be the asymptote for us.

00:41 So there is no y intercept.

00:47 And the next one, determine the horizontal and the vertical asymptote.

00:50 So for the vertical asymptote, so what we have to do, we have to just set the denominator equal to zero.

01:00 So from here, x is equal to zero.

01:02 Is the vertical asymptote and for the horizontal asymptote we have to just take the limit of the function when x goes to positive or negative infinity which is e to the infinity over two divided by infinity which is equal to infinity over infinity if we take the derivative of this function by using the lopital rule so which is limit when x goes to infinity which is one over two times e to the x over 2 and divided by 1.

01:36 If you plug in here, which is 1 over 2 times e to the power infinity, which is 1 over 2 times infinity, which is infinity 2.

01:43 So that means there is no horizontal asymptote that we have for this function.

01:50 And the next part, we have to just find the first interval of the function, the first interval.

01:57 So first of all, the x values are both in the denominator and the numerator.

02:02 So we have to just apply the division rule here.

02:06 So the derivative of this function equal to the derivative of the top which is one half times a to power x over two times the denominator minus the derivative of the denominator which is one times the top and divided by the square of the denominator.

02:22 So for the critical points and also for the rise and the faults of the function we have to just set this equation equal to zero.

02:30 So from here at the top if i do the cross multiplication that up we have the common factor which is e2 power x over 2 which is x over 2 minus 1 that is equal to 0.

02:41 If i set each equation, each factor equal to 0, we got x over 2 minus 1, which is equal to 0.

02:49 So from here, x would be 2.

02:51 So we have one critical point here.

02:54 And for the fall and the rise of the function, we have to just make a sign table for the first derivative of the function.

03:01 So the leading coefficient for the function, which is plus...

Question

Please give Ace some feedback