Determine the value of kkk such that f(x)={ekx−1xx<03x+k2x≥0f(x) = \begin{cases} \frac{e^{kx}-1}{x} & x < 0 \\ 3x + k^2 & x \ge 0 \end{cases}f(x)={xekx−13x+k2x<0x≥0 is continuous at x=0x=0x=0.
k=1k=1k=1
k=2k=2k=2
k=1k=1k=1 or k=−2k=-2k=−2
k=0k=0k=0