False

The curve $f'(x) = \dfrac 1{\sqrt{x^3+2}}$ is shown in the graph. And $f(x)$ is given by the area under the curve and above the $x$ -axis between 0 and $x$ .

• A: $\displaystyle f(0) = \int_0^0 \frac 1{\sqrt{t^3+1}} dt = 0$ . True .
• B: Since $\dfrac 1{\sqrt{x^3+1}}$ is continuous for $x \ge 0$ , $f$ is continuous at all $x \ge 0$ . True .
• C: Area under the curve from $x=0$ to $x=1$ is positive, hence $f(1) > 0$ . True .
• D: $f'(x) = \dfrac 1{\sqrt{x^3+2}} \implies f'(1) = \dfrac 1{\sqrt{1^3+2}} = \dfrac 1{\sqrt 3}$ . True .
• E: $\displaystyle f(-1) = \int_0^{-1} \frac 1{\sqrt{t^3+2}} dt = - \int_{-1}^0 \frac 1{\sqrt{t^3+2}} dt$ . Since from the graph $\displaystyle \int_{-1}^0 \frac 1{\sqrt{t^3+2}} dt > 0$ , $f(-1) < 0$ . Therefore, $f(-1) >0$ is FALSE .