So far I gained to calculating the derivative 5x^4 +2" and second derivative 20x^3"" yet then I got stuck here. Many thanks ahead of time because that the aid :D The an initial derivative is offered by

#f"(x) = 5x^4 +2#

Recall that horizontal tangents will occur when #f"(x) = 0#.

#0 = 5x^4 +2#

#-2/5 = x^4#

#x= root(4)(-2/5)#

This is undefined, as such the role has no horizontal tangents.

To find the tangent with the minimum slope, we seek to minimization #5x^4 + 2#. You"re right that the derivative that this will be #20x^3#. This will have a crucial point in ~ #x= 0#, which will certainly be a minimum since #f""(-1) = -# and also #f""(1)= +# and also #f""(0) = 0#.

Therefore, the value of #x# that offers the the smallest tangent slope will be #x =0# and also the slope will certainly be #2#.

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Hopefully this helps! LM
Feb 17, 2018

#5x^4+2>=2#smallest slope possible: #2#

Explanation:

the gradient of a horizontal heat is #0#.

the derivative of a role can be supplied to find the gradient of a line tangent come the graph.

you have offered the derivative that the duty #y = x^5+2x#; that is #5x^4+2#.

all genuine numbers have squares that are either positive, or #0#.

#x^4 = (x^2)^2#

the square of any kind of positive number is likewise positive.the square that #0# is #0#.

this way that #x^4# must constantly be #0# or above.

#5x^4 + 2# must thus be #2# or above.

since no tangent to the graph #y=x^5+2x# can have a gradient equal to #0#, there deserve to be no horizontal tangents.

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the smallest slope feasible can be discovered by calculating the value of #x# when the second derivative is #0#.

(note the all gradients #5x^4 +2#, for any kind of real worth of #x#, room non-negative.)

the second derivative offered here is #20x^3#.

#20x^3 = 0#

#x^3 = 0#

#x = root3(0) = 0#

then the worth of #x# that makes the second derivative #0# deserve to be substituted into the equation because that the first derivative.

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#x = 0#

#5x^4 + 2 = (5*0^4) + 2#

#= (5*0) + 2#

#= 0 + 2#

#= 2#

the the smallest value feasible for the derivative, and also therefore the smallest feasible value because that the slope of the tangent, is #2#.