Question:
Given f(x) = e^(Kx)+x and f’0 =-8 . Find k
Answer:
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Given f(x)=e^(kx) +x and f'(0)= -8
Let us derivate both sides with respect to “x”
f'(x)= d/dx ( e^kx) + d/dx(x)
f'(x) = ke^(kx) +1
when x=0
f'(0) = ke^(k*0) +1
f'(0)= ke^0 +1
f'(0)= k+1
But we are given that f'(0) = -8 in the original question itself.
hence
-8=k+1
k=-8-1 =-9
K=-9
