The number of distinct real roots of the equation x^7-7x-2=0 is

Step-by-step idea

1. Polynomial behaviour at the ends
   For very large negative $x$, the term $x^7$ dominates, so
   $\lim_{x \to -\infty} (x^7 - 7x - 2) = -\infty$
   For very large positive $x$, $\lim_{x \to +\infty} (x^7 - 7x - 2) = +\infty$
   Thus the curve comes from $-\infty$ on the left and shoots to $+\infty$ on the right. At least one root is guaranteed.

2. Critical points (where slope is zero)
   Find the derivative: $f'(x) = 7x^6 - 7 = 7\bigl(x^6 - 1\bigr) = 7\bigl(x^3 - 1\bigr)\bigl(x^3 + 1\bigr)$
   Set $f'(x)=0$: $x^3 = 1 \;\;\Longrightarrow\;\; x = 1$
   $x^3 = -1 \;\;\Longrightarrow\;\; x = -1$
   So there are only two turning points: $x=-1$ and $x=1$.

3. Nature of these points
   Because $f'(x)$ changes sign:
   • Increasing for $x<-1$
   • Decreasing for $-1<x<1$
   • Increasing for $x>1$

   Hence $x=-1$ is a local maximum, and $x=1$ is a local minimum.

4. Evaluate $f(x)$ at the critical points
   $f(-1) = (-1)^7 - 7(-1) - 2 = -1 + 7 - 2 = 4$
   $f(1) = 1^7 - 7(1) - 2 = 1 - 7 - 2 = -8$

5. Sign changes in each interval
   • Interval $(-\infty, -1)$: $f$ starts at $-\infty$ and climbs to $+4$
   → must cross the axis once.
   • Interval $(-1, 1)$: $f$ goes from $+4$ down to $-8$
   → must cross the axis once.
   • Interval $(1, \infty)$: $f$ rises from $-8$ up to $+\infty$
   → must cross the axis once.

6. Maximum possible roots per interval
   Because the function is monotonic in each interval (no extra turning points), it can cross at most once in each. We already showed it does cross each time.

Hence, total distinct real roots = 3.

What is the question?

We have a big, bendy line (a 7th–degree polynomial):

$x^7 - 7x - 2 = 0$

We only want to know how many times this line touches the ground (the x-axis). Each touch is a real root.

How do we find that out?

1. Look at the slopes: Where does the line change from going up to going down?
2. Check the height at those change points: Is the line above or below the ground?
3. See the pattern: If it starts below ground, rises above, then dips below, then rises again, you’ll get three touches.

That is exactly what happens here, so we get 3 real roots.

Complete working

Let $f(x)=x^7-7x-2$.

1. Derivative

Setting $f'(x)=0$ gives $x = -1,\; 1.$ So these are the only critical points; the function is monotonic in each of the three intervals $(-\infty,-1]$, $[-1,1]$, and $[1,\infty)$.

2. Function values at critical points

3. Sign analysis

• As $x\to -\infty$, $f(x)\to -\infty$; at $x=-1$, $f=+4$
⇒ one root in $(-\infty,-1)$.

• From $x=-1$ ($f=+4$) to $x=1$ ($f=-8$) and function strictly decreasing
⇒ one root in $(-1,1)$.

• At $x=1$, $f=-8$ and as $x\to+\infty$, $f(x)\to+\infty$ with function strictly increasing
⇒ one root in $(1,\infty)$.

Adding up: $1+1+1 = 3$ distinct real roots.

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