Answer
The function $~~f(x) = x^{101}+x^{51}+x+1~~$ has neither a local maximum nor a local minimum.
Work Step by Step
Note that the function is continuous and differentiable for all real numbers.
We can find the values of $x$ when $f'(x) = 0$:
$f(x) = x^{101}+x^{51}+x+1$
$f'(x) = 101x^{100}+51x^{50}+1 = 0$
Let $y = x^{50}$. Then:
$101x^{100}+51x^{50}+1 = 101y^2+51y+1 = 0$
We can use the quadratic formula to find the values of $y$:
$y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$y = \frac{-51 \pm \sqrt{(51)^2-4(101)(1)}}{(2)(101)}$
$y = \frac{-51 \pm \sqrt{2197}}{202}$
$y = -0.020435, -0.4845$
Since $y = x^{50}$, we can set up two equations:
$x^{50} = -0.020435$
$x^{50} = -0.4845$
However, since $x^{50} \geq 0$ for all real numbers, these two equations have no solutions.
Then there are no values of $x$ such that $f'(x) = 0$.
Since the function is continuous and differentiable for all real numbers, the function has neither a local maximum nor a local minimum.
