How many real roots does the equation $x^{5}-7 x+c=0$ have in the interval $[-1,1]$ ?
A) at most five real roots
B) no real roots
C) at most one real root
D) at most two real roots
E) at most three real roots

Question

Quizplus · Accepted Answer

Since $x^5-7x$ has alternating signs on the interval $[-1,1]$, by the intermediate value theorem, the number of real roots is equal to the number of times the graph of $y=c$ intersects the graph of $y=x^5-7x$ in the interval $[-1,1]$. Since $x^5-7x$ has only one real root in $[-1,1]$, namely 0, there can be at most one real root of $x^5-7x+c$ in $[-1,1]$. Therefore, the answer is $\boxed{	extbf{(C)}}$.

How Many Real Roots Does the Equation x5−7x+c=0x^{5}-7 x+c=0x5−7x+c=0 Have in the Interval

How Many Real Roots Does the Equation $x^{5}-7 x+c=0$ Have in the Interval