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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