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

  • $\begingroup$ Hint - Note that $t^2$ is always positive, so to get a value smaller than $108$, there must exist at least one $t$ s.t. $10t^6 - 24 t^5 + 15t^4 <0 \iff 10t^2 - 24t + 15 <0$. Now what can you say about this quadratic inequality? $\endgroup$ – stochasticboy321 Dec 27 '16 at 5:58
  • $\begingroup$ If you don't want to take concept of derivative then you can draw graph for that function and check minimum value. $\endgroup$ – Fawad Dec 27 '16 at 6:01

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 $t$?


Let's look at $f(t)-108$. One has


Now let's have a look at $g(t)-40$


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.


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 $t$


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