Answer
$T (x)=x^4+3x-1+.......$; converges for all $x$ with convergence interval $(-\infty,\infty)$.
Work Step by Step
The Taylor`s Series can be represented as: : $T(x)=\Sigma_{n = 0}^{\infty} \dfrac{f^n (c)}{n!}(x-c)^n$
If we consider $c=0$, then Taylor`s Series, $T(x)$ is also called as Maclaurin Series and the Maclaurin series for $f(x)$ is as given below:
$f(x)=f(0)+f^{\prime}(0) x+\dfrac{f^{\prime \prime}(0) x^2}{2!}+\dfrac{f^{\prime \prime \prime}(0) x^3}{3!}+\dfrac{f^{\prime \prime \prime \prime}(0) x^4}{4!}+..............$
Here, we have
$f(x)=x^4+3x-1 \implies f(0)=-1\\ f'(x)=4x^3+3 \implies f'(0)=3\\f''(x)=12x^2 \implies f''(0)=0\\f'''(x)=24x \implies f'''(0)=0\\f''''(0)=24 \implies f^{iv}(2)=24$
Thus, we have the Taylor`s Series: $T (x)=1+3(x-0)+\dfrac{0}{2!}(x-0)^2+\dfrac{0}{3!}(x-0)^3+..............\\=x^4+3x-1+.......$; converges for all $x$ with convergence interval $(-\infty,\infty)$.
