minimum value of $f(t) = 10t^6-24t^5+15t^4+40t^2+108$ without derivative minimum value of $f(t) = 10t^6-24t^5+15t^4+40t^2+108$ without derivative
for $t\leq 0$ value of function $f(t)\geq 108$
i wan,t be able to proceed after that ,could some help me with this
 A: Note that 
\begin{align}
f(t) &= 10t^6-24t^5+15t^4+40t^2+108
\\
&= 10t^4\left(t^2-\dfrac{12}{5}t+\dfrac{3}{2}\right)+40t^2+108
\\
&= 10t^4\left(t^2-\dfrac{12}{5}t+\dfrac{36}{25}+\dfrac{3}{50}\right)+40t^2+108
\\
&= 10t^4\left(\left(t-\dfrac{6}{5}\right)^2+\dfrac{3}{50}\right)+40t^2+108.
\end{align}
Now, can you show that $f(t) \ge 108$ for all real $t$?
A: Let's look at $f(t)-108$. One has
$$f(t)-108=t^2\left(10t^4-24t^3+15t^2+40\right)=t^2g(t)$$
Now let's have a look at $g(t)-40$
$$g(t)-40=t^2\left(10t^2-24t+15\right)$$
The quadratic factor of $g(t)$ has discriminant $\delta=144-150=-6$. So it has the sign of its leading coefficient $+10$ and so $g(t)\gt 40$. And this means $f(t)-108\geq 0$ i.e. $108$ is the minimum we're looking for.
A: for $t\leq 0\;,$ then $f(t) = 10t^6-24t^5+15t^4+40t^2+108\geq 108$
for $t\geq 0\;,$ then $f(t) = t^2(10t^4+15t^2-24t^3)+108\geq t^2(2\sqrt{10t^4\cdot 15t^2}-24t^3)+108$
$\Rightarrow f(t) =t^2(10t^4+15t^2-24t^3)+108\geq t^2(10\sqrt{6}-24t^3)+108\geq 108$
$\Rightarrow f(t) = 10t^6-24t^5+15t^4+40t^2+108\geq 108$ for all real values of $t$
