Question

Let $f\left(x\right)=[\begin{array}{ccc}\frac{e^{3x}-1}{x}& if& x\ne 0\\ 3& if& x=0\end{array}$ then $f^{\prime }\left(0\right)$, is?

• A. $-9$
• B. $9$
• C. $9/2$
• D. Non existent

Answer 1

Answer: (C)

$9/2$

$\text{Given}\text{that}$

$f\left(x\right)=$ $\frac{e^{3x}-1}{x},x\ne 0$

$3,x=0$

Using formula

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+......+......$

Now,

$f\left(x\right)=\frac{1+3x+\frac{{\left(3x\right)}^2}{2!}+\frac{{\left(3x\right)}^3}{3!}+......-1}{x}$

$=\frac{3x+\frac{9x^2}{2!}+\frac{27x^3}{3!}}{x}$

$f\left(x\right)=3+\frac{9x}{2!}+\frac{27}{3!}{x}^2+......$

On differntaning and we get.

$f^{\prime }\left(x\right)=0+\frac{9}{2!}+\frac{54}{3!}x+......$

$Takingx=0$

So,

$f^{\prime }\left(0\right)=\frac{9}{2!}$

$f^{\prime }\left(0\right)=\frac{9}{2}$

$Hence,Thisistheanswer.$