dalhousie calculus is killing me!! "Let ƒ(x)=3x3 + √x-2..."

[tnigsnevin]

dalhousie calculus is killing me!! "Let ƒ(x)=3x3 + √x-2..."

1. Let ƒ(x)=3x³ + √x-2
a. Find an interval where the function ƒ has one root.
b. Use Rolle’s theorem to show that the function ƒ has exactly one root.

Plz help???

 

[pka]

1. Let ƒ(x)=3x³ + √x-2
a. Find an interval where the function ƒ has one root.
b. Use Rolle’s theorem to show that the function ƒ has exactly one root.

Prove that \(\displaystyle f(0)<0~\&~f(1)>0\) AND \(\displaystyle \forall x\in(0,1)[f'(x)>0]\).

That answers both questions.
BTW: I have never liked authors who use Rolle's theorem for these.

 

[stapel]

1. Let ƒ(x)=3x³ + √x-2

As posted, the function is this:

. . . . .\(\displaystyle \mbox{i. }\, f(x)\, =\, 3x^3\, +\, \sqrt{\strut x\,}\, -\, 2\)

Was this what you meant, or were grouping symbols omitted, and the function is actually as follows?

. . . . .\(\displaystyle \mbox{ii. }\, f(x)\, =\, 3x^3\, +\, \sqrt{\strut x\, -\, 2\,}\)

a. Find an interval where the function ƒ has one root.

Hint: How do "roots" relate the x-intercepts? What is the value of y at any x-intercept? What can be said about the sign of y on either side of an x-intercept where the graph passes through the x-axis? How might this relate to finding one x-value at which your function is positive, and another x-value at which your function is negative? What can be said about the (continuous) function between these two x-values?

b. Use Rolle’s theorem to show that the function ƒ has exactly one root.

What is your book's statement of "Rolle's Theorem"? What have you done with this information?

Please be complete. Thank you!

 

[pka]

As posted, the function is this:

. . . . .\(\displaystyle \mbox{i. }\, f(x)\, =\, 3x^3\, +\, \sqrt{\strut x\,}\, -\, 2\)

Was this what you meant, or were grouping symbols omitted, and the function is actually as follows?

. . . . .\(\displaystyle \mbox{ii. }\, f(x)\, =\, 3x^3\, +\, \sqrt{\strut x\, -\, 2\,}\)

Just an observation: \(\displaystyle f(x)\, =\, 3x^3\, +\, \sqrt{\strut x\, -\, 2\,}\) has no zeros because its domain is \(\displaystyle x\ge 2\) and the function is increasing.