Let f (x) = x · ekx (K ≠ 0) ((ekx) ′ = kekx) (1) find the tangent equation of curve y = f (x) at point (0, f (0)); (2) find the monotone interval of function f (x)

Let f (x) = x · ekx (K ≠ 0) ((ekx) ′ = kekx) (1) find the tangent equation of curve y = f (x) at point (0, f (0)); (2) find the monotone interval of function f (x)

(1) F ′ (x) = ekx + kxekx = (1 + KX) ekx (x ∈ R), and f ′ (0) = 1, the tangent slope is 1, and f (0) = 0, the tangent equation of the curve y = f (x) at the point (0, f (0)) is X-Y = 0. (2) f ′ (x) = (KX + 1) ekx (x ∈ K), Let f ′ (x) = 0, then x = - 1K. ① if k > 0, when x ∈ (- ∞, - 1K), f ′ (x) < 0, f (x) monotonically decreases; if x ∈ (- ∞, - 1K), f ′ (x) < 0, f (x) monotonically decreases When ∈ (- 1K, + ∞), f ′ (x) > 0, f (x) increases monotonically. ② if K < 0, when x ∈ (- ∞, - 1K), f ′ (x) > 0, f (x) increases monotonically; when x ∈ (- 1K, + ∞), f ′ (x) < 0, f (x) decreases monotonically. In conclusion, when k > 0, the monotone decreasing interval of F (x) is (- ∞, - 1K), and the monotone increasing interval is (- 1K, + ∞); when k < 0, f (x) decreases monotonically (x) The monotone increasing interval is (- ∞, - 1K), and the monotone decreasing interval is (- 1K, + ∞);