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(d) Differentiate each of the following functions and find the value of the derivative when x = 1:

(i) f(x) = 2 + x 3 (7 ln(x) − 9),

(ii) g(x) = 1 + 2x 2 e x 1 + x 2 ,

(iii) h(x) = √ 4 + x 4 .

 

(e) Let f(x) = x 3 − 2x 2 − 4x + 5.

(i) Find f 0 (x) and f 00(x).

(ii) Determine the critical points of f.

(iii) Show that x = −2/3 is a local maximum of f.

(iv) Show that x = 2 is a local minimum of f.

 Jul 28, 2016
 #1
avatar+23245 
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Let me try part e:     f(x)  =  x3 - 2x2 - 4x + 5

 

i)     f'(x)  =  3x2 - 4x - 4

       f''(x)  =  6x - 4

 

ii)     The critical points occur where the first derivate either does not exist or it equals 0.

        Since the first derivative is always defined, we need to find where it equals 0.

               f'(x)  =  3x2 - 4x - 4  =  0     --->     3x2 - 4x - 4  =  0     --->     (3x + 2)(x - 2)  =  0

                                                         --->      x  =  -2/3     or     x = 2           <---     the critical points

 

iii)     A local maximum occurs where the first derivative equals 0 and the second derivative is less than 0.

              f'(-2/3)  =  0

              f''(-2/3)  =  6(-2/3) - 4  =  -8         <---     Therefore, a local maximum occurs where  x = -2/3.

 

iv)     A local minimum occurs where the first derivative equals 0 and the second derivative is greater than 0.

              f'(2)  =  0

              f''(2)  =  6(2) - 4  =  8          <--     Therefore, a local minimum occurs where x = 2.

 Jul 28, 2016
 #2
avatar+23245 
0

Let me try part d     ---     but I'm guessing at what you mean ....

 

i)     f(x)  =  2 + x3(7·ln(x) - 9)          Using the product rule:

       f'(x)  =  2 + x3(7/x) + 3x2(7·ln(x) - 9)

       f'(x)  =  2 + 7x2 + 3x2(7·ln(x) - 9)

       f'(1)  =  2 + 7·12 + 3·12(7·ln(1) - 9

              --->   f'(1)  =  2 + 7 + 3(7·0 - 9)   =   2 + 7 - 27   =  -18

 

ii)    g(x)  =  1 + 2x2e+ x2                        [Is this right?]

       g'(x)  =  0 + 2x2ex + 4x·e+ 2x

       g'(x)  =   2x2ex + 4x·e+ 2x

              --->   g'(1)  =  2·12·e1 + 4·1·e1 + 2·1  =  2e + 4e + 2  =  6e + 2

 

iii)   h(x)  =  sqrt(4) + x4                 [Is this right?]

       h'(x)  =  0 + 4x3

       h'(x)  =  4x3

              --->   h'(1)  =  4·13  =  4

 Jul 28, 2016

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