Question

Examine the function f(x) = x^3 - 5x^2 +8x-4 for maxima and minima.
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Answer 1

Answer:

Step-by-step explanation:

F'(x)=0 is minima value at x=2

And

F'(x)=4/27 is maxima value at n =4/3

Answer 2

Given,

the function f(x) = $x^{3} - 5x^{2} +8x - 4$                       (1)

To Find,

find maxima and minima for the given equation.

Solution,

differentiating equation (1) w.r.t 'x'

y'(x) = $3x^{2} -10x-8$                                                 (2)

to find the critical numbers,set y'(x) = 0

$3x^{2} -10x-8$ = 0

(3x+2)(x-4)=0

the critical numbers are x = $-\frac{2}{3}$ , x= 4

now, differentiating equation (2) w.r.t 'x'

y''(x) = 6x-10

y''($-\frac{2}{3}$) = -14 <0

so, y($-\frac{2}{3}$) = 157/27 corresponds to a relative maximum value of the function.

y''(4) = 14>0

so, y(4) = -45 corresponds to relative minimum value of the function.

for inflection point solve y''(x)=0

6x - 10 = 0

x=5/3

x< 5/3, y''(x) <0, so the graph is concave downward on (-inf,5/3)

x>5/3, y''>0,so the graph is concave upward on (5/3,inf)

the graph has an inflection point at (5/3 , -529/27)

Hence the maxima is -2/3 and the minima is 4.