Question

Consider the function below.

f(x) =

x^2 / ( x - 8 ) ^2

(a) Find the vertical and horizontal asymptotes.

x=

y=

(b) Find the interval where the function is increasing. (Enter your answer using interval notation.)

Find the interval where the function is decreasing. (Enter your answer using interval notation.)

(c) Find the local minimum value.

(d) Find the inflection point.
(x, y) =

e :Find the interval where the function is concave up. (Enter your answer using interval notation.)

f: Find the intervals where the function is concave down. (Enter your answer using interval notation.)

Answer 1

f(u) = n² (2-02 same (a) vertical asymptote is given by (2-0)2 =0 n = 8 since degree of numkrator is as that of denominator Hence Horizonal asymptote y= = 1 = leading coefficient of numerator leading cofficient of denominator (b) fino se² (n-07 d 28² _ x² d du dni f(n)= (2-0) (2-0)2 (x-014 (using quotient rule of differentiati zul2-822222(2-0) (n-074 2n (n -07[x In-814

. rogon - 16n(n-do (n-014 foul is increasons film) > 0 - 16x(n-0) > 0 aluodo co osac Interval is (0,8) fin) is decreasi decreasing when fln) Ro -lon(n- oco >o r(u-8) a <o, a > 8 (-0,0 ) U (8,00) ® for Local extremum f'() co - 1674-0) (n-a3

(4-873 af 6-164) -G-16912-033 Now f'(4)= [(n=0)3 7² = -1610-8)3 + 1 Gux3172-012 (3-876 16(n-8} +421 (0-0)2 In-dio = -1661–812 [~-8 -32] In-86 = +16 (21+8) (n-0) f"(0) 16x8 (-876 Hence there is local minimum > at u 0 fro) o local minimum value for inflection pönts f"lns=0 16 (2n to) 50-) / 5-4 Tn-o) f1-4)= t4,2 16 (-4-82² Inflection point (-4,/9) 144

@ find is concare up when f"(x) > 0 16(2018) (n-016 > 0 La u >- (-4,00) © fen) is concave down f'! () a <-4 (-00,-4)