
Math
We need to take the derivative of the function first in order to identify the critical points or critical numbers, which we can use for the endpoints of our intervals. For the right side we need to...

Math
This function is defined everywhere except x=0 and is differentiable. To determine where it is increasing or decreasing, compute the derivative: `y'(x)=1  9/x^2.` It is positive in `(oo, 3) uuu...

Math
`f(x)=cosx+tanx` Mean value theorem can be applied, 1. if f is continuous on the closed interval `[a,b]` , 2. if f is differentiable on the open interval (a,b) 3. there is a number c in (a,b) such...

Math
(1) `f(x)=x/(x+1) ` is a rational function. It has a vertical asymptote at x=1, and a horizontal asymptote of y=1. On the interval [1/2,2] the graph is increasing towards the limiting value of...

Math
f(pi) = pi, f(pi) = pi. The secant line goes through the points `(pi, pi)` and `(pi, pi)` . The equation is obviously y=x, its slope is 1. Let's find points where f'(x)=1: f'(x) = 1  2cos(x)....

Math
a) the graph is here: https://www.desmos.com/calculator/kdzzqcpt3q b) the endpoints of the graph are (0, 0) and (9, 3). The slope of the secant line is 1/3, the equation is y=x/3. c)...

Math
Given `f(x)=x^42x^3+x^2 ` on the interval [0,6]: (1) This is a quartic polynomial with positive leading coefficient so its end behavior is the same as a parabola opening up. (2) f(0)=0 and...

Math
Consider x=1 and x=0. For x=1 f(x)<0 and for x=0 f(x)>0. By Intermediate Value Theorem there is at least one root of f at the interval (1, 0). Let's suppose that there are two roots. Then...

Math
`f(x)=2x^5+7x1` `f(0)=2*0^5+7*01=1` `f(1)=2*1^5+7*11=8` So,f(0) is negative and f(1) is positive. Since f(x) is continuous, by the Intermediate value theorem there is a number c between 0 and 1...

Math
Consider f(x)=3x+1sinx. This function is continuous and infinitely differentiable on `RR`. `f(0)=1gt0` and `f(pi)=13pilt0.` By the Intermediate Value Theorem there is at least one `c in (pi,...

Math
You need to evaluate the derivative of the function `f(x) = 2x  2  cos x` , such that: f'(x) = 2 + sin x You need to use Rolle's theorem, so you need to find the roots of the equation 2 + sin x =...

Math
The function whose derivative is always 0 is a constant function. Given that at x=2 f(x)=5 we make a conclusion that this constant is 5. The answer: f(x)=5. (no, I cannot use 120 words here)

Math
You need to notice that if the derivative is a constant function, then the primitive is a linear function, such that: `f(x) = ax + b` Differentiating f(x) yields: `f'(x) = a` You need to set equal...

Math
You need to notice that if the derivative is a linear function, then the primitive is a quadratic function, such that: `f(x) = ax^2 + bx + c` Differentiating f(x) yields: `f'(x) = 2ax + b` You need...

Math
You need to notice that if the derivative is a linear function, then the primitive is a quadratic function, such that: `f(x) = ax^2 + bx + c` Differentiating f(x) yields: `f'(x) = 2ax + b` You need...

Math
Rolle's Theorem requires f to be defined and continuous on the given closed interval, differentiable on the open interval and values of f on ends to be equal. Here all conditions are met...

Math
NO. This function isn't defined on entire interval (the point where it isn't defined is pi/2). Actually, not only Rolle's Theorem isn't applicable but also its conclusion doesn't hold: there is no...

Math
The Rolle's theorem is applicable to the given function, only if the function is continuous and differentiable over the interval, and f(a) = f(b). Since all trigonometric functions are continuous...

Math
Rolle's theorem cannot be applied because the function is not differentiable over the whole interval `(1,1).` More specifically the function is not differentiable at zero. Graph of the function...

Math
You need to notice that the given function is continuous on [0,1] and differentiable on (0,1), since it is a polynomial function. You need to verify if f(0)=f(1), hence, you need to evaluate the...

Math
NO, Rolle's Theorem isn't applicable because the function doesn't has equal values at the endpoints: f(1/4) = 1/4  tan(pi/4) = 1/4  1 = 3/4 while f(1/4) = 1/4  tan(pi/4) = 1/4 + 1 = 3/4.

Math
The given function is continuous and differentiable over the given interval, as all trigonometric functions are. For Rolle's Theorem to be applied, you also need to test if `f(1) = f(0).` `f(1) =...

Math
The mean value theorem is applicable to the given function, since it is a polynomial function. All polynomial functions are continuous and differentiable on R, hence, the given function is...

Math
The mean value theorem is applicable to the given function, since it is a polynomial function. All polynomial functions are continuous and differentiable on R, hence, the given function is...

Math
Yes, it can. The function is continuous on [1, 1] and is differentiable on (1, 1). Here a=1 and b=1. f(a) = f(1) = 3 and f(b) = f(1) = 3. So `(f(b)f(a))/(ba) = 6/2 = 3.` f'(x) = 3x^2 +...

Math
The mean value theorem may be applied to the given function since all polynomial functions are continuous and differentiable on R, hence, the given function is continuous on [0,2] and...

Math
The mean value theorem is applicable to the given function, since it is a polynomial function. All polynomial functions are continuous and differentiable on R, hence, the given function is...

Math
The mean value theorem is applicable to the given function, since it is a polynomial function. All polynomial functions are continuous and differentiable on R, hence, the given function is...

Math
NO, it isn't. The function is continuous on [1, 3] and is differentiable on (1, 3) but the point where 2x+1=0, or x=1/2 in [1, 3]. To the left of this point f(x)=2x1, to the right f(x)=2x+1,...

Math
The mean value theorem is applicable to the given function, since it is a polynomial function. All polynomial functions are continuous and differentiable on R, hence, the given function is...

Math
Yes, it can. The function f is contionuous on ` [0, pi ]` and is differentiable on `(0, pi )`, which are all the requirements for the Mean Value Theorem. Therefore there is at least one point c on...

Math
Rolle's Theorem requires the function to be continuous on the closed interval [a, b]. But this function isn't. The problem point is x=0. Aclually, f(x) isn't even defined at x=0, and because...

Math
The Rolle's theorem is applicable to the given function, only if the function is continuous and differentiable over the interval, and f(a) = f(b). Since all trigonometric functions are continuous...

Math
This function satisfies some conditions of Rolle's Theorem: it is continuous on [0, 2], differentiable almost everywhere on (0, 2) and f(0) = f(2) (=0). But there is one point where f isn't...

Math
You need to notice that the given function is continuous on [1,1] and differentiable on (1,1), since it is a polynomial function. You need to verify if `f(1)=f(1), ` hence, you need to evaluate...

Math
The xintercepts are roots of f(x)=0. There are only two of those: 1 and 2. The point between them where f'(x)=0 exists by Rolle's Theorem and we can find it: f'(x)=2x1, it is zero only at x=1/2.

Math
You need to find the two x intercepts of the function, hence, you need to solve for x the equation f(x) = 0, such that: `f(x) = x^2 + 6x = 0` You need to factor out x, such that: `x(x + 6) = 0...

Math
You need to find the two x intercepts of the function, hence, you need to solve for x the equation f(x) = 0, such that: `f(x) = x*sqrt(x+4) = 0` `x*sqrt(x+4) = 0 x = 0` `sqrt(x+4) = 0 => x + 4 =...

Math
Hello! f(x) = 0 for x=0 and x=1 (obvious). f is continuous on [1, 0] and is differentiable on (1, 0) (as an elementary function). So we can aplly Rolle's Theorem and conclude that there is at...

Math
The Rolle's theorem is applicable to the given function, only if the function is continuous and differentiable over the interval, and f(a) = f(b). Since all polynomial functions are continuous and...

Math
Given: `f(x)=x^28x+5,[2,6].` Rolle's Theorem can be applied because the function f(x) is a continuous polynomial on the closed interval [2,6] and differentiable on the open interval (2,6)., and...

Math
Yes it can. The function f is continuous on [1, 3] and differentiable on (1, 3) (as an elementary function, or even as polynomial) and f(1) = f(3) (=0). All conditions are met. The value of c is...

Math
The Rolle's theorem is applicable to the given function, only if the function is continuous and differentiable over the interval, and f(a) = f(b). Since all polynomial functions are continuous and...

Math
Given: f(x)=x^(2/3)1,[8,8]. Rolle's Theorem does not apply to the function f(x) on the given interval because all 3 conditions of Rolle's Theorem will not be met. The function f(x) is continuous...

Math
Given f(x)=3x3 on the interval [0,6]: To apply Rolle's theorem the function must be continuous on the closed interval [a,b], differentiable on the open interval (a,b), and f(a)=f(b). f is...

Math
The Rolle's theorem is applicable to the given function, only if the function is continuous and differentiable over the interval, and f(a) = f(b). Since all polynomial functions are continuous and...

Math
The Rolle's theorem is applicable to the given function, only if the function is continuous and differentiable over the interval, and f(a) = f(b). Since all polynomial functions are continuous and...

Math
Yes, it can. Function f is continuous on `[0, 2pi]` and is differentiable on` (0, 2pi)` . Also, `f(0) = f(2pi)` (both =0). There are all conditions of Rolle's Theorem. Because of this there is at...

Math
The Rolle's theorem is applicable to the given function, only if the function is continuous and differentiable over the interval, and f(a) = f(b). Since all trigonometric functions are continuous...

Math
The Rolle's theorem is applicable to the given function, only if the function is continuous and differentiable over the interval, and f(a) = f(b). Since all polynomial functions are continuous and...