Please answer all for number 2?

Please show full working out

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1 Answer
Oct 27, 2017

I have corrected part b.

Explanation:

a) #(x+h)^3 = (x+h)(x+h)(x+h) = (x+h)(x^2+2xh+h^2) = x^3+3x^2h+3xh^2+h^3#

b) For this #f(x)# the derivative is #f'(x) = 6x^2-6# Not #6x-6#

#f'(x) = lim_(hrarr0)(f(x+h)-f(x))/h#

#= lim_(hrarr0)([2(x+h)^3-6(x+h)]-[2x^3-6x])/h#

# = lim_(hrarr0)([2(x^3+3x^2h+3xh^2+h^3)-6(x+h)]-[2x^3-6x])/h#

# = lim_(hrarr0)(2x^3+6x^2h+6xh^2+2h^3-6x-6h-2x^3+6x)/h#

# = lim_(hrarr0)(6x^2h+6xh^2+2h^3-6h)/h#

# = lim_(hrarr0)(6x^2+6xh+2h^2-6)#

# = 6x^2-6#

c) #f'(x) > 0# on #(-oo,-1)# and on #(1,oo)# and
#f'(x) < 0# on #(-1,1)#.
Therefore, #f# is decreasing on #(-1,1)#.

So #p# and #q# can be any numbers in #[-1,1]# with #p < q#

d) #f''(x) = 12x#

e) #f# is concave up (the graph of #f# is concave up) where #f''(x) > 0#. So the graph of #f# is concave up on #(0,oo)#.

Assuming that the question wants the maximal open interval on which #f# is concave up, the answer should be #(0,oo)#.