#maxima and minima of a function but there's more to it
north pewter
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https://math.stackexchange.com/questions/4733384/if-fx-fracx55-fracx44x3-frackx2...
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north pewter
If $f(x)=\frac{x^5}5+\frac{x^4}4+x^3+\frac{kx^2}2+x$ be a real valued function. Find the maximum value of $k^2$ for which $f(x)$ is increasing $\forall x \in R$.
I differentiated the function, and set up the inequality where the derivative is greater than $0$. Then I rearranged the terms to get $$k<\frac1x +x^3+x^2+3x$$
To find the maxima and minima of the right hand expression, I took the derivative and setting it equal to $0$, obtained:
$$3x^4+2x^3+3x^2-1=0$$
- How do I factorize this?
- How do I solve this equation(if factorizing is not the optimal way)
- Now about the graph which I checked using an online calculator, I am getting that $f'(x)=\frac1x +x^3+x^2+3x$
has the range $[- \infty,-3.322]\cup[3.862,\infty ]$. If $k$ is
always less than $f'(x)$, then doesn't that mean that $k$ ranges
from $[-\infty, \infty]$? So $k^2$ ranges from $[0, \infty]$? But
the answer is given $0.322^2=11.03...$. How am I interpreting the
inequality wrong?
hidden blazeBOT
strange_quark_
north pewter
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