Question

Which of the following is (are) correct about the function $y=x^4-\frac{4x^3}{3}$ ?

• A. f is decreasing in (– ∞, 1]
• B. f is strictly increasing in (– ∞, 1]
• C. f is increasing in [1, ∞).
• D. f is decreasing in (1, ∞).

Answer 1

Answer: (A), (C)

f is increasing in [1, ∞).

$\mathrm{Given}\mathrm{the}\mathrm{function},y=x^4-\frac{4x^3}{3}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=4x^3-4x^2=4x^2\left(x-1\right)\mathrm{Now},\frac{\mathrm{dy}}{\mathrm{dx}}=0\Rightarrow 4x^2\left(x-1\right)=0\Rightarrow x=0\mathrm{or}1$
Since f ′ (x) < 0 ∀ ∈ x (– ∞, 0) ∪ (0, 1) and f is continuous in (– ∞, 0] and [0, 1]. Therefore f is decreasing in (– ∞, 1] and f is increasing in [1, ∞).