Chapter 4, Problem 54RE

To determine

Tofind:the function using its first derivative.

Answer to Problem 54RE

Thefunction using its first derivative $\text{f}\left(\text{u}\right)\text{=}\frac{{\text{u}}^{\text{2}}}{\text{2}}{\text{+2u}}^{\frac{\text{1}}{\text{2}}}+\frac{1}{2}.$

Explanation of Solution

Given:

  $\text{<msup><mo>f</mo><mo>′</mo></msup>}\left(\text{u}\right)\text{=}\frac{{\text{u}}^{\text{2}}\text{+}\sqrt{\text{u}}}{\text{u}}\text{, f}\left(\text{1}\right)\text{=3}\text{.}$ .

Concept used:

Antiderivative=integration

  ${\displaystyle \int \text{<msup><mo>f</mo><mo>′</mo></msup>}\left(\text{t}\right)}\text{dt=f}\left(\text{t}\right)$ . Or ${\displaystyle \int \frac{\text{dy}}{\text{dt}}}\text{dt=y}$ .

Since, the multiplication of two operators ${\displaystyle \int \times \frac{\text{d}}{\text{dt}}}=1$ .

Integration formula:

  ${\displaystyle \int {\text{x}}^{\text{n}}}\text{dx=}\frac{{\text{x}}^{\text{n+1}}}{\text{n+1}}$ .

Calculation:

  $\text{<msup><mo>f</mo><mo>′</mo></msup>}\left(\text{u}\right)\text{=}\frac{{\text{u}}^{\text{2}}\text{+}\sqrt{\text{u}}}{\text{u}}\text{, f}\left(\text{1}\right)\text{=3}\text{.}$ .

  ${\displaystyle \int f^{\prime }\left(u\right)}\text{du=}{\displaystyle \int \frac{{\text{u}}^{\text{2}}\text{+}\sqrt{\text{u}}}{\text{u}}du}\text{, f}\left(\text{1}\right)\text{=3}\text{.}$ .

  $\text{f}\left(\text{u}\right)\text{=}{\displaystyle \int \text{udu}}\text{+}{\displaystyle \int \frac{\text{1}}{\sqrt{\text{u}}}}\text{du, f}\left(\text{1}\right)\text{=3}\text{.}$

  $\text{f}\left(\text{u}\right)\text{=}\frac{u^2}{2}\text{+}{\displaystyle \int u^{\frac{-1}{2}}}\text{du, f}\left(\text{1}\right)\text{=3}\text{.}$

  $\text{f}\left(\text{u}\right)\text{=}\frac{{\text{u}}^{\text{2}}}{\text{2}}\text{+}\frac{{\text{u}}^{\frac{\text{1}}{\text{2}}}}{\frac{\text{1}}{\text{2}}}\text{+C, f}\left(\text{1}\right)\text{=3}\text{.}$

  $\text{f}\left(\text{u}\right)\text{=}\frac{{\text{u}}^{\text{2}}}{\text{2}}{\text{+2u}}^{\frac{\text{1}}{\text{2}}}\text{+C, f}\left(\text{1}\right)\text{=3}\text{.}$

  $\text{3=}\frac{1^{\text{2}}}{\text{2}}\text{+2}{\left(1\right)}^{\frac{\text{1}}{\text{2}}}\text{+C}\text{.}$

  $\text{3=}\frac{1}{2}\text{+2+C}\text{.}$

  $\frac{1}{2}=C$ .

  $\text{f}\left(\text{u}\right)\text{=}\frac{{\text{u}}^{\text{2}}}{\text{2}}{\text{+2u}}^{\frac{\text{1}}{\text{2}}}+\frac{1}{2}.$

Hence the function using its first derivative $\text{f}\left(\text{u}\right)\text{=}\frac{{\text{u}}^{\text{2}}}{\text{2}}{\text{+2u}}^{\frac{\text{1}}{\text{2}}}+\frac{1}{2}.$