Free math worksheets

- Check continuity on the full closed interval first. - Then check differentiability only at interior points. - A corner and a point where a function is undefined cause different theorem hypotheses to fail.

1. \(f(x)=\sqrt{x}\) is continuous on \([1,4]\) and differentiable on \((1,4)\), so the theorem applies. 2. \(g(x)=|x|\) is continuous on \([-1,1]\) but is not differentiable at \(x=0\), so the theorem is not guaranteed to apply. 3. \(h(x)=\frac{1}{x-2}\) is not continuous on \([1,3]\) because it is undefined at \(x=2\), so the theorem is not guaranteed to apply.

a) Yes. b) No; differentiability fails at \(x=0\). c) No; continuity fails at \(x=2\).

- Start with the rate of change determined by the two endpoint values. - Ask what the theorem says about that rate somewhere inside the interval. - Separate the information needed for the theorem from information needed to solve an equation explicitly.

1. The average rate of change is \(\frac{17-5}{8-2}=2\). 2. Differentiability on \((2,8)\) together with continuity on \([2,8]\) satisfies the hypotheses of the Mean Value Theorem. 3. Therefore, for some \(c\in(2,8)\), \(f'(c)=2\).

a) \(f'(c)=2\) for at least one \(c\in(2,8)\). b) The Mean Value Theorem uses the endpoint values and the continuity/differentiability hypotheses; an explicit formula is not required.

- Find the change in height per unit of elapsed time over the whole interval. - Interpret the derivative of height in this context. - State why the theorem's hypotheses are satisfied.

1. The average vertical velocity is \(\frac{300\,\text{ft}}{20\,\text{s}}=15\,\text{ft/s}\). 2. The height function is differentiable and therefore continuous on the time interval. 3. By the Mean Value Theorem, at some interior time the instantaneous vertical velocity equals the average vertical velocity, so it is \(15\,\text{ft/s}\).

At some time strictly inside the \(20\,\text{s}\) interval, the elevator's instantaneous vertical velocity is \(15\,\text{ft/s}\).

- Compare the theorem's hypotheses with the type of function shown. - The relevant line slope comes from the two endpoints of the interval. - Parallel lines have equal slopes.

1. The polynomial \(f\) is continuous on \([0,4]\) and differentiable on \((0,4)\), so the Mean Value Theorem applies. 2. \(f(0)=3\) and \(f(4)=11\), so the secant slope is \(\frac{11-3}{4}=2\). 3. Since \(f'(x)=2x-2\), solve \(2c-2=2\), giving \(c=2\).

a) The theorem applies. b) The tangent is parallel to the secant at \(x=2\).

- Separate conditions on the function from conditions on its derivative. - Recall exactly which interval is closed and which is open in the theorem statement. - Ask whether the theorem says anything about continuity of the derivative itself.

1. The statement is false. 2. The Mean Value Theorem requires \(f\) to be continuous on the closed interval and differentiable on the open interval. 3. Continuity of \(f'\) is not one of the hypotheses. A derivative may exist throughout the open interval without itself being continuous.

False. The theorem requires continuity of \(f\) on \([a,b]\) and differentiability of \(f\) on \((a,b)\); it does not require \(f'\) to be continuous.

- Use only the endpoint rows to find the interval's overall rate of change. - Interpret the derivative of position in context. - The theorem guarantees an interior time even though that time is not listed in the table.

1. The average velocity is \(\frac{133-85}{20-12}=\frac{48}{8}=6\,\text{m/s}\). 2. Since \(s\) is differentiable, it is continuous on the interval. 3. By the Mean Value Theorem, there is some \(c\in(12,20)\) with \(s'(c)=6\,\text{m/s}\).

At some time \(c\in(12,20)\), the runner's instantaneous velocity is \(6\,\text{m/s}\).

- Use the change in position and elapsed time to find the interval's average velocity. - Connect that average velocity to an instantaneous derivative value using the theorem. - Convert the guaranteed velocity to speed before comparing it with the speed limit.

1. The average velocity over the interval is \(\frac{120\,\text{mi}}{1.5\,\text{h}}=80\,\text{mph}\). 2. The position function is continuous on the closed time interval and differentiable in its interior, so the Mean Value Theorem gives an interior time \(c\) with \(s'(c)=80\,\text{mph}\). 3. Since this velocity is positive, the instantaneous speed at that time is \(|s'(c)|=80\,\text{mph}\), which exceeds the \(70\,\text{mph}\) speed limit.

The instantaneous speed was \(80\,\text{mph}\) at some time during the trip, so the car exceeded a \(70\,\text{mph}\) speed limit.

- Check the theorem's conditions separately on the closed interval and the open interval. - Compare the slope across the whole interval with a tangent-line slope inside the interval. - Translate “parallel” into a condition involving two slopes.

1. The polynomial \(f\) is continuous on \([0,2]\) and differentiable on \((0,2)\), so the Mean Value Theorem applies. 2. \(f(0)=0\) and \(f(2)=2\), so the secant slope is \(\frac{2-0}{2-0}=1\). 3. Since \(f'(x)=3x^2-3\), solve \(3c^2-3=1\). Thus \(c^2=\frac{4}{3}\), and the value in \((0,2)\) is \(c=\frac{2\sqrt{3}}{3}\).

a) The theorem applies because \(f\) is continuous on \([0,2]\) and differentiable on \((0,2)\). b) Secant slope: \(1\). c) \(c=\frac{2\sqrt{3}}{3}\).

- Relate total change across the interval to one derivative value inside it. - Use the derivative bounds to bound that total change. - After finding a candidate range, check whether a constant-slope function can realize every value in it.

1. By the Mean Value Theorem, there is some \(c\in(1,4)\) such that \(f(4)-f(1)=f'(c)(4-1)\). 2. Since \(2\le f'(c)\le5\), \(6\le f(4)-7\le15\). 3. Therefore, \(13\le f(4)\le22\). 4. Conversely, for any \(L\in[13,22]\), the linear function \(f(x)=7+\frac{L-7}{3}(x-1)\) has derivative \(\frac{L-7}{3}\in[2,5]\) and satisfies \(f(4)=L\). Thus every value in the interval is possible.

\(f(4)\in[13,22]\).

- Compute the slope determined by the two endpoint values. - Solve for every interior point whose derivative equals that slope. - Distinguish what the theorem guarantees from what the algebra happens to show for this function.

1. The polynomial is continuous on \([0,3]\) and differentiable on \((0,3)\). 2. \(f(0)=0\) and \(f(3)=6\), so the average rate of change is \(2\). 3. Since \(f'(x)=3x^2-6x+2\), solve \(3c^2-6c+2=2\), giving \(3c(c-2)=0\). The value in \((0,3)\) is \(c=2\). 4. The Mean Value Theorem guarantees at least one such value; uniqueness is not part of its conclusion. In this particular function, solving the derivative equation shows that the only interior value is \(2\).

a) \(c=2\). b) No. The theorem guarantees existence, not uniqueness; uniqueness here follows from solving the derivative equation.

- Do not start with the endpoints only; choose two arbitrary points in the interval. - Apply the theorem on the smaller interval determined by those two points. - Use the derivative information to determine the corresponding secant slope.

1. Choose any two points \(u<v\) in \([a,b]\). 2. The Mean Value Theorem applied on \([u,v]\) gives a point \(c\in(u,v)\) such that \(\frac{f(v)-f(u)}{v-u}=f'(c)\). 3. Since \(f'(c)=0\), \(f(v)-f(u)=0\), so \(f(v)=f(u)\). 4. Because the two points were arbitrary, \(f\) has the same value everywhere on \([a,b]\), so it is constant.

For arbitrary \(u<v\), the Mean Value Theorem gives \(\frac{f(v)-f(u)}{v-u}=0\), hence \(f(v)=f(u)\). Therefore, \(f\) is constant on \([a,b]\).

- Determine the slope forced by the two endpoint values. - Ask what derivative value the theorem then guarantees. - Compare that required value with the proposed derivative bound.

1. The average rate of change from \(0\) to \(4\) is \(\frac{15-2}{4}=\frac{13}{4}=3.25\). 2. By the Mean Value Theorem, some \(c\in(0,4)\) must satisfy \(f'(c)=3.25\). 3. This contradicts \(f'(x)\le3\) everywhere on the open interval. Therefore, no such function exists.

The claim is impossible because the Mean Value Theorem would require \(f'(c)=3.25\) for some \(c\in(0,4)\), contradicting \(f'(x)\le3\).

- For a piecewise function, check the joining point separately. - Continuity and differentiability require two different comparisons at that point. - After finding the interval's average slope, solve on each derivative piece.

1. Each piece is continuous and differentiable on its own interval. At \(x=1\), the left and right function values are both \(1\), so \(f\) is continuous there. 2. The left derivative at \(1\) is \(2\), and the right derivative is also \(2\), so \(f\) is differentiable at \(1\). Therefore, the Mean Value Theorem applies on \([0,3]\). 3. \(f(0)=0\) and \(f(3)=5\), so the average rate of change is \(\frac{5}{3}\). 4. On \((0,1)\), \(f'(x)=2x\), so \(2c=\frac{5}{3}\) gives \(c=\frac{5}{6}\). On \((1,3)\), \(f'(x)=2\), so there is no additional solution.

a) The theorem applies; the pieces match in both value and derivative at \(x=1\). b) \(c=\frac{5}{6}\).

- Break the full interval at the middle point. - Compare the endpoint values on each smaller interval. - The two theorem applications produce points in disjoint open intervals.

1. Apply the Mean Value Theorem on \([a,b]\). Since \(f(a)=f(b)\), the secant slope is \(0\), so there is \(u\in(a,b)\) with \(f'(u)=0\). 2. Apply the Mean Value Theorem on \([b,c]\). Since \(f(b)=f(c)\), there is \(v\in(b,c)\) with \(f'(v)=0\). 3. The intervals \((a,b)\) and \((b,c)\) are disjoint, so \(u\ne v\). Thus there are at least two distinct zeros of \(f'\).

There is a point \(u\in(a,b)\) and a distinct point \(v\in(b,c)\) such that \(f'(u)=f'(v)=0\).

- Compare the two functions by considering their difference. - What is the derivative of that difference? - Use the theorem to determine what a function with zero derivative can do across an interval.

1. Define \(h(x)=f(x)-g(x)\). Then \(h\) is continuous on \([0,5]\), differentiable on \((0,5)\), and \(h'(x)=f'(x)-g'(x)=0\). 2. For any \(x\in(0,5]\), apply the Mean Value Theorem to \(h\) on \([0,x]\). There is some \(c\in(0,x)\) such that \(h(x)-h(0)=h'(c)x=0\), so \(h(x)=h(0)\). 3. Since \(h(0)=f(0)-g(0)=0\), \(h(x)=0\) for every \(x\in(0,5]\); it is also \(0\) at \(x=0\). Therefore, \(f(x)=g(x)\) throughout \([0,5]\).

\(f(x)=g(x)\) for every \(x\in[0,5]\).