Find the coordinates of the turning point of the curve y=8x+(1/2x^2) and determine whether this point is a maximum or minimum point?

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Answer 1 · 2018-01-18T22:29:46

#(-8.-32)#; minimum point

#y = 8x + (1/2x^2)#

#f(x) = 1/2x^2 + 8x#

#(Deltay)/(Deltax) (x^2/2) = 2 * x/2 = x#

#(Deltay)/(Deltax) (8x) = 8#

#(Deltay)/(Deltax) (1/2x^2 + 8x) = x+8#

at a turning point, #(Deltay)/(Deltax) = 0#.

#(Deltay)/(Deltax) = x+8 = 0#

#x = 0-8 = -8#

either side of #x=-8:#

when #x= -7.9999#, #(Deltay)/(Deltax) = 0.0001#

when #x = -8#, #(Deltay)/(Deltax) = 0#

when #x = -8.0001#, #(Deltay)/(Deltax) = -0.0001#

to the left of #x=-8#, the gradient is negative.
at #x-8#, the gradient is #0#.
to the right of #x=-8#, the gradient is positive.

this shows that #(-8, y)# is a minimum point.

graph{8x + (1/2)x^2 [-80, 80, -40, 40]}

#y = 8x + (1/2)x^2#

#y = -64 + (64/2)#

#=-64 + 32#

#=-32#

#f(-8) = -32)#