Question

Let f"(x)=2x+1/2 x^(-1/2),f'(1)=2 and f(1)=3. Find f(x).

Answer 1

It is given that :-

$\tt \large \:f"(x) = 2x + \dfrac{1}{2} {x}^{ \frac{ - 1}{2} } \\ \\ \tt \large f'(1) = 2 \\ \\ \tt \large \: f(1) = 3$

Now , first we will find the f'(x).

To find f'(x) we have to integrate f''(x).

$\tt \longrightarrow \displaystyle \int \tt \: f"(x)dx = \displaystyle \int \tt(2x + \dfrac{1}{2} {x}^{ \frac{ - 1}{2} })dx \\ \\ \tt \longrightarrow \tt \: f '(x)= \tt { \: {x}^{2} } + {x}^{ \frac{1}{2} } + c$

Here, c is constant.

To find the value of C , put x = 1

f'(1) = 2 [ given ]

$\tt \longrightarrow \tt \: f '(1)= \tt { \: {1}^{2} } + {1}^{ \frac{1}{2} } + c \\ \\ \tt \longrightarrow \tt \: 2 = \tt { \: {1}^{2} } + {1}^{ \frac{1}{2} } + c\\ \\ \tt \longrightarrow \tt \: 2 = \tt { \: {1}} + {1}+ c\\ \\ \tt \longrightarrow \tt \: 0 = c$

$\tt \therefore f '(x)= \tt { \: {x}^{2} } + {x}^{ \frac{1}{2} } + 0 \\ \\ \tt \rightarrow f '(x)= \tt { \: {x}^{2} } + {x}^{ \frac{1}{2} }$

Now we will find f(x) by integrating f'(x) :-

$\tt \longrightarrow \displaystyle \tt \int f '(x) \: dx= \displaystyle \tt \int ({ \: {x}^{2} } + {x}^{ \frac{1}{2} })dx \\ \\ \tt \longrightarrow \tt f (x) = \displaystyle \tt { \dfrac{ {x}^{3} }{3} } + \dfrac{2 \: {x}^{ \frac{2}{3}}}{3} + k$

Here, K is constant.

To find the value of K , put x = 1

f (1) = 3 [ given ]

$\longrightarrow \tt f (1) = \displaystyle \tt { \dfrac{ {1}^{3} }{3} } + \dfrac{2 \: {(1)}^{ \frac{2}{3}}}{3} + k \\ \\ \longrightarrow \tt f (1) = \displaystyle \tt { \dfrac{ {1} }{3} } + \dfrac{2 }{3} + k\\ \\ \longrightarrow \tt 3 = \displaystyle \tt \dfrac{1 + 2 }{3} + k\\ \\ \longrightarrow \tt 3 = \displaystyle \tt 1 + k\\ \\ \longrightarrow \tt 2 = \displaystyle \tt k$

$\therefore \: \tt f (x) = \displaystyle \tt { \dfrac{ {x}^{3} }{3} } + \dfrac{2 \: {x}^{ \frac{2}{3}}}{3} + 2$