Question

7. Show that the equation f(x) = x^3 + 3x^2 - 9x + 7 = 0 has a solution for some x is E(-6; -5).
Apply Newton’s method with an initial guess x0 = -5 to find x2.

8. Find the intervals of increase and decrease of the function x2e^-2x.

9. Sketch the graph of the curve y = x3 + 3x2 - 9x + 7. Be sure to find the intervals of increase,
decrease and constant concavity and all local extremes and inflection points and all intercepts.
If you don’t know the intercept(s) exactly, use Newton’s method to get it accurarte to one
tenth.

Answer 1

Solu, Giren us. equation fees x + 3x² - gx +7=0 Newton's Method have In a fren) Xntia f'(on) 2881.2 differenkieling eq en watex f'exp= 3x²+63 -9 By newton's Method. Xo = 5 $(5) 5'+3 (51%-965) +7 = 162 fles) 3(5)² +6(5) ag = 96 - 5 X = Xo- Xo - (15) 1'(no) Xy - 5- 162 - 3.3125 96 Now ah X, = -3.3125 (-3.3125) -(-3.3125) +3 (-3.3125)°-9(-3.3125)+7 f(0) = -26.24 48-3.3125) : 3 (-3.3125+ 6(-3.3125) - 9 f'(x2) = 4.035

By newhon methods U2 X, f(xi) 5'(4) 1-26.24 CE X = -3.3125 4.035 2 = -3.3125 +6.50 X = 3.1875 of. -0) -27, - 2 - 2x - 2x 2x - е obson 87 The intervals of increar and decrease f(x) = x?é?* differentiating with r.h. x dy x² e 2: (-2) - 2x de Cat (2) dy - - 2x e dx > dy - exe9 [x --1 dy da le 2010 - 2x e ie x=0 dx putting as - O - 2x & lx=1

8 1 to Intervals signs decrease tit (-0,0) (0,1) (1,0) Increase Intrete decrease ti 11) Ctit Increase 111) g Given us. x + 3x -9N+7 for a differentiating both side w.o.o. I fill) = 3x'+63-9 (x) =o emisia put f'le) = 3x?+63-9-0 naisina x +2% -3,- 0 2?+ 3x - 1-3-0 (x+3)-1(1+3) = 0 (X +3 )(x-1)=0 X=-3 X-1 oo 18 signs Intervals (-00,-3) (-3 11 ) (10) increasing, (dy dx) decreasing inereasing + I +

and for extreme values. Again differentiating eq (1) wird a alors p"(m)6x +6 me # At x = -3 le - 34 1"(-3) = -18+6 =-12 (negative), le maxima So, Maximum value f(-5) = (-3) + 3(-3) -9(-3) +7 is a maximum value mam biz sod At xal f(1) = 6+6 – 12 ( positive) in minima 0.00 minimum value - - x + x 40:1-360 -961+1 .4 is the minimum value soi