3.3 given $f(t)=\frac{2}{3}t^{2}-\frac{1}{t^{4}}+\frac{3}{t}-sqrt{t}+25$, determine $f(t)$ and evaluate…

Explanation:

Step1: Rewrite the function

Rewrite $f(t)$ as $f(t)=\frac{2}{3}t^{2}-t^{- 4}+3t^{-1}-t^{\frac{1}{2}} + 25$.

Step2: Find the first - derivative $f'(t)$

Using the power rule $\frac{d}{dt}(t^n)=nt^{n - 1}$, we have: $f'(t)=\frac{2}{3}\times2t-(-4)t^{-5}+3\times(-1)t^{-2}-\frac{1}{2}t^{-\frac{1}{2}}+0=\frac{4}{3}t + 4t^{-5}-3t^{-2}-\frac{1}{2}t^{-\frac{1}{2}}$.

Step3: Find the second - derivative $f''(t)$

Again, using the power rule: $f''(t)=\frac{4}{3}+4\times(-5)t^{-6}-3\times(-2)t^{-3}-\frac{1}{2}\times(-\frac{1}{2})t^{-\frac{3}{2}}=\frac{4}{3}-20t^{-6}+6t^{-3}+\frac{1}{4}t^{-\frac{3}{2}}$.

Step4: Evaluate $f''(t)$ at $t = 1$

Substitute $t = 1$ into $f''(t)$: $f''(1)=\frac{4}{3}-20\times1^{-6}+6\times1^{-3}+\frac{1}{4}\times1^{-\frac{3}{2}}=\frac{4}{3}-20 + 6+\frac{1}{4}$. First, find a common denominator, which is 12. $\frac{4}{3}=\frac{16}{12}$, $-20=-\frac{240}{12}$, $6=\frac{72}{12}$, $\frac{1}{4}=\frac{3}{12}$. $f''(1)=\frac{16-240 + 72+3}{12}=\frac{91 - 240}{12}=-\frac{149}{12}$.

Answer:

$f''(t)=\frac{4}{3}-20t^{-6}+6t^{-3}+\frac{1}{4}t^{-\frac{3}{2}}$, $f''(1)=-\frac{149}{12}$