f(x)=e4x+2e3x−ex−6 f′(x)=4e4x+6e3x−ex =ex(4e3x+6e2x−1) Let g(x)=ex h(x)=4e3x+6e2x−1 g(x)>0,∀x∈R h′(x)=12e3x+12e2x=12e2x(ex+1) h′(x)>0,∀x∈R h(x) is an increasing function. Minimum value of h(x) will be when x→−∞ at [h(x)]min=−1 and [h(x)]max=∞ f′(x)=g(x)⋅h(x) Now, h(x) is an increasing function and h(x) varies from −1 to +∞ . So, this implies that h(x) cuts the X -axis at one point and which further implies f(x) changes its sign only at one point. Let's say at x=α f(x)=e4x+2e3x−ex−6 When, x→−∞;f(x)→−6 x→+∞;f(x)→+∞