1. **State the problem:** Find the second derivative $f''(t)$ of the function $$f(t) = 5t^2 - \frac{1}{t^3} + 3t - \sqrt{t} + 1$$ and then evaluate it at $t=1$. 2. **Rewrite the function with exponents:** $$f(t) = 5t^2 - t^{-3} + 3t - t^{\frac{1}{2}} + 1$$ 3. **Find the first derivative $f'(t)$ using power rule:** - Derivative of $5t^2$ is $10t$ - Derivative of $-t^{-3}$ is $3t^{-4}$ - Derivative of $3t$ is $3$ - Derivative of $-t^{\frac{1}{2}}$ is $-\frac{1}{2}t^{-\frac{1}{2}}$ - Derivative of constant $1$ is $0$ So, $$f'(t) = 10t + 3t^{-4} + 3 - \frac{1}{2}t^{-\frac{1}{2}}$$ 4. **Find the second derivative $f''(t)$:** - Derivative of $10t$ is $10$ - Derivative of $3t^{-4}$ is $-12t^{-5}$ - Derivative of $3$ is $0$ - Derivative of $-\frac{1}{2}t^{-\frac{1}{2}}$ is $\frac{1}{4}t^{-\frac{3}{2}}$ So, $$f''(t) = 10 - 12t^{-5} + \frac{1}{4}t^{-\frac{3}{2}}$$ 5. **Evaluate $f''(t)$ at $t=1$:** $$f''(1) = 10 - 12(1)^{-5} + \frac{1}{4}(1)^{-\frac{3}{2}} = 10 - 12 + \frac{1}{4} = -2 + \frac{1}{4} = -\frac{7}{4}$$ **Final answer:** $$f''(1) = -\frac{7}{4}$$