How do you find the slope of #f(x) = 0.5x^2+6x+7.5# at #x=-6# by first principles?

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Answer 1 · 2015-07-04T09:56:40

Evaluate the slope at #x = -6# by looking at the limit of

#(f(x+epsilon) - f(x))/epsilon# as #epsilon->0# at that point.

#f(x+epsilon) = 0.5(x+epsilon)^2+6(x+epsilon)+7.5#

#=0.5x^2+6x+7.5+epsilon x+0.5epsilon^2+6epsilon#

So:

#(f(x+epsilon) - f(x))/((x+epsilon)-x)#

#=(epsilon x+0.5epsilon^2+6epsilon)/epsilon#

#=x+6+0.5epsilon#

So when #x=-6#

#(f(x+epsilon) - f(x))/((x+epsilon)-x) = 0.5epsilon -> 0# as #epsilon -> 0#