Determine if the function [tex]f(x)=\sqrt[5]{x^4}[/tex] is differentiable at [tex]x=0[/tex]. If not, identify why.
a.) [tex]f(x)=\sqrt[5]{x^4}[/tex] is not differentiable at [tex]x=0[/tex] because [tex]f(0)[/tex] is not defined.
b.) [tex]f(x)=\sqrt[5]{x^4}[/tex] is not differentiable at [tex]x=0[/tex] because [tex]\lim _{x \rightarrow 0} \sqrt[5]{x^4}[/tex] does not exist.
c.) [tex]f(x)=\sqrt[5]{x^4}[/tex] is not differentiable at [tex]x=0[/tex] because [tex]f^{\prime}(0)[/tex] is not defined.
d.) [tex]f(x)=\sqrt[5]{x^4}[/tex] is differentiable at [tex]x=0[/tex].
Answer (3)
The first one is of order 5, so it has either 1, 3 or 5 real roots (unless any coefficent was complex). Proof complete :) The other one, if it has a solution, it must be in [-1;1]. Because it only gives positive results the solution is further restricted to [0;1]. Because the cosine function is continuous and strictly decreasing on this interval, the difference of x and it's cosine will shrink up to some point within the interval where it gets to 0 (the solution) and then flips sign (the cosine gets less than the number), further decreasing until the end of the interval.
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The polynomial x 5 − x 2 + 2 x + 3 = 0 has at least one real root due to the Intermediate Value Theorem, as it changes signs between evaluated points. Similarly, the equation x = cos ( x ) has at least one solution because both x and cos ( x ) intersect within the interval [ − 1 , 1 ] . Thus, both equations have at least one real root or solution.
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