Show that the equation x^3 - 14x + 3 = 0 has three solutions in the interval [-4

Question

Show that the equation x^3 - 14x + 3 = 0 has three solutions in the interval [-4, 4]. Let f(x) be equal to the left side of the equation. Where do the solutions to the equation exist? Solution exist at values of x_1 and x_2 if f(x_1) f(x_2). Solution exist at values of x_1 and x_2 if 0 lessthanorequalto f(x_1) lessthanorequalto f(x_2) or 0 greaterthanorequalto f(x_1) greaterthanorequalto f(x_2) Solution exist between values of x_1 and x_2 if f(x_1) lessthanorequalto 0 and f(x_2) greaterthanorequalto 0 or f(x_1) greaterthanorequalto 0 and f(x_2) lessthanorequalto 0. Solution exist between values of x_1 and x_2 if f(x_1) lessthanorequalto 0 and f(x_2) lessthanorequalto 0 or f(x_1) greaterthanorequalto 0 and f(x_2) greaterthanorequalto 0.

Explanation / Answer

x3-14x+3=0

Interval is [-4,4]

Lets take some points in the interval. Lets take 0,1,4,-4

f(0)=3 >0

f(1)= -10<0

f(4)= 11>0

f(-4)= -5<0

As we see that f(0)>0 and f(1)<0, so according to Intermediate value theorem, there must exist c between 0 and 1 such that f(c)=0

Similarly f(1)<0 and f(0)>0, therefore their must exist c between 1 and 4 such that f(c)=0

And f(-4)<0 , f(0)>0, there must exist c between -4 and 0 such that f(c)=0

So we see that there ust be at least 3 such solutions

And the correct option is C

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.