# Find the relative extrema of $3x^4−36x^3+140x+200$

Find the relative extrema of the function:

$$3x^4−36x^3+140x+200$$

So I found the 1st derivative:

$$12x^3-108x^2+140$$

Then I got lost from here because I don't think it can be factorized, I'm trying to get the critical points then find the critical values. I'm sorry if my equations are not too good, I still don't know the most suitable equation editors for this site

• Are you sure it's $140^{x}$? Feb 8, 2021 at 13:42
• In either case, according to WolframAlpha, the relative extrema are not very 'nice'. Feb 8, 2021 at 13:46
• That's where the confusion came from but it's what my maths lecturer gave us as assignment Feb 8, 2021 at 13:46
• The derivative of $140^x$ is not $140$ but rather $\ln(140)\,140^x$. Feb 8, 2021 at 13:48
• Meta-cheating, instructors don't normally inflict such difficult problems, when the obvious point of such an expression is for you to realize the significance of the first and second derivatives. The place for you to start is to re-read and respond to the comment of Joshua Wang. Feb 8, 2021 at 14:02

The derivative of $$140^x$$ is not $$140$$. Keep in mind that there are four distinct formulae for evaluating the derivative of a number raised to another number: \begin{align} \frac{d}{dx}(a^b) &= 0 \\ \frac{d}{dx}(x^a) &= ax^{a-1} \\ \frac{d}{dx}(a^x) &= a^x \log a \\ \frac{d}{dx}(x^x) &= x^x \left(\log x + 1\right) \end{align} The first two formulae should be familiar to you: the derivative of a constant is zero, and the derivative of $$x^a$$, where $$a$$ is a constant, simply involves the application of the power rule. The last two might not be so familiar. But it is the third, not the second formula that applies here: $$\frac{d}{dx}(140^x) = 140^x \log(140) \, .$$ If you use this formula, then should be able to find the extrema of the function is question by setting the first derivative equal to $$0$$.

• Wow, I didn't know the derivative will not be 140, thanks for this. Though I'm still unable to find the extrema of the function because I'm not aligned to the 140x log(140) function Feb 8, 2021 at 15:18
• @Chinez I think you should speak to your teacher about this. As people have already noted in the comments, this problem is unlikely to have a 'nice' solution. You have an $x$ as the base of an exponential function on one term and as the power of an exponential function in another. This means that the answer cannot be solved using 'elementary methods', and has to be approximated using numerical methods.
– Joe
Feb 8, 2021 at 15:42
• Well I made a mistake, it's 140x, not 140^x sorry Feb 8, 2021 at 15:51
• @Chinez In that case, how are you finding this problem? Have you got the solution now?
– Joe
Feb 9, 2021 at 8:59
• @Chinez: Good news! And good luck with your studies. For me, calculus is the most beautiful area of mathematics.
– Joe
Feb 12, 2021 at 9:12