Free Single Variable Calculus Course Materials Study Resources

To calculate the work required to pump the liquid, we use the integral $W = \int dW. 1. **Establish a Coordinate System**: Let y represent the vertical distance from the top rim of the tank (y=0) down to the bottom (y=4). 2. **Analyze a Slice**: Consider a horizontal slice of water at depth y$ with thickness $dy. The slice is a circle with radius r. Using the Pythagorean theorem for the hemispherical profile, r^2 + y^2 = R^2$, so $r^2 = 16 - y^2. 3. **Force of the Slice**: The volume of this slice is dV = \pi r^2 dy = \pi(16 - y^2) dy. The weight (force) required to lift it is dF = \rho g dV = 1000 \cdot 9.8 \cdot \pi(16 - y^2) dy = 9800\pi(16 - y^2) dy. 4. **Work for the Slice**: To lift this slice to the top rim, it must travel a distance d = y$. Thus, $dW = y \cdot dF = 9800\pi(16y - y^3) dy$. 5. Integration: Summing the work for all slices from $y=0$ to $y=4$: $W = \int_{0}^{4} 9800\pi(16y - y^3) dy = 9800\pi [8y^2 - \frac{1}{4}y^4]_0^4$. Evaluating at the boundaries: $W = 9800\pi [8(16) - \frac{1}{4}(256)] = 9800\pi [128 - 64] = 9800\pi(64) = 627,200\pi\text{ J}$. Answer: C

To find the derivative of the polynomial $f(x) = 4x^3 + 2x^2 + 5$, we apply the Power Rule and Sum Rule term-by-term. First, applying the Power Rule ($d/dx[x^n] = nx^{n-1}$) and Constant Multiple Rule to the first term: $d/dx(4x^3) = 4(3x^2) = 12x^2$. Second, applying it to the second term: $d/dx(2x^2) = 2(2x^1) = 4x$. Third, the derivative of the constant term 5 is 0. Summing these results gives $f'(x) = 12x^2 + 4x$. The correct option is B.

To evaluate the limit $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$, follow these steps:

1. Check for Indeterminate Form: Substitute $x = 3 into the expression. We get \frac{3^2 - 9}{3 - 3} = \frac{0}{0}$. This is an indeterminate form, which means we must use algebraic techniques to simplify the expression.

2. Factor the Numerator: The numerator $x^2 - 9 is a difference of squares and can be factored as (x - 3)(x + 3)$.

3. Simplify the Expression: Rewrite the limit with the factored form: $\lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3}$ As long as $x \neq 3, we can cancel the common factor (x - 3)$ from the numerator and denominator: $\lim_{x \to 3} (x + 3)$

4. Evaluate the Limit: Now, substitute $x = 3$ into the simplified expression: $3 + 3 = 6$

Therefore, the value of the limit is 6.

To determine where the function is concave upward, we need to analyze the sign of the second derivative, $f''(x)$. The graph of $f is concave upward on intervals where f''(x) \ge 0$.

Step 1: Find the first and second derivatives. $f(x) = x^4 + ax^3 + 6x^2$ $f'(x) = 4x^3 + 3ax^2 + 12x$ $f''(x) = 12x^2 + 6ax + 12$

Step 2: Establish the condition for global concavity. We require $f to be concave upward for all real x$, which means $f''(x) \ge 0$ for all $x$. Since $f''(x) = 12x^2 + 6ax + 12$ is a quadratic with a positive leading coefficient ($12 > 0), the parabola opens upward. For the values to remain non-negative everywhere, the quadratic must have at most one real root (it must not cross the x-axis into negative values). This occurs when the discriminant, $\Delta$$, is less than or equal to zero.

Step 3: Calculate and analyze the discriminant. The discriminant of a quadratic $Ax^2 + Bx + C$ is $$\Delta = B^2 - 4$AC$. Here, $A = 12$, $B = 6a$, and $C = 12$.

$\Delta = (6a)^2 - 4(12)(12)$ $\Delta = 36a^2 - 576$

Set \Delta$ \le 0$: 36a^2 - 576 \le 0$$36a^2 \le 576$Divide by 36:$a^2 \le \frac{576}{36}$$a^2 \le 16$$

Taking the square root of both sides (considering the absolute value): $|a| \le 4$ $-4 \le a \le 4$

Conclusion: The values of $a for which the function is concave upward everywhere are in the interval **[-4, 4]$**.

Analysis of Distractors:

• Option B: This interval corresponds to $a^2 \ge 16$, which is where the function would have two inflection points and regions of downward concavity.
• Option C: This result ($a^2 \le 4) might arise from failing to multiply by 4 in the discriminant formula (B^2 - AC \le 0$).
• Option D: This excludes the endpoints. While concavity is strictly positive on $(-4, 4), the standard definition allows for inflection points where concavity doesn't change sign (like y=x^4$ at $x=0), or simply the boundary condition$\Delta$\le 0$ which includes $a = \pm 4$ where $f''(x)$ touches 0 but stays non-negative.

1. Identify the substitution: Let $u = \tan(x). The differential is then du = \sec^2(x) dx$, which matches the term in the integral.
2. Change the limits of integration:
   • When $x = 0$, $u = \tan(0) = 0$.
   • When $x = \frac{\pi}{4}$, $u = \tan(\frac{\pi}{4}) = 1$.
3. Rewrite the integral in terms of $u$: $\int_{0}^{1} u \, du$
4. Evaluate the integral using the power rule: $\left[ \frac{1}{2}u^2 \right]_{0}^{1} = \frac{1}{2}(1)^2 - \frac{1}{2}(0)^2 = \frac{1}{2}$

Final Answer: $\frac{1}{2}$

To determine the possible range of values for $f(5), we apply the Mean Value Theorem (MVT). The MVT states that for a function f$ continuous on $[a, b]$ and differentiable on $(a, b), there exists at least one point c \in (a, b)$ such that:

$f'(c) = \frac{f(b) - f(a)}{b - a}$

In this problem, $a = 1$, $b = 5$, and $f(1) = 7. Rearranging the MVT formula to solve for f(5)$ gives:

$f(5) = f(1) + f'(c)(5 - 1)$ $f(5) = 7 + 4f'(c)$

We are given the bounds for the derivative: $-2 \le f'(x) \le 3$. Since $c is a point in the interval (1, 5)$, $f'(c)$ must also satisfy these bounds:

1. Lower Bound: To find the minimum possible value for $f(5), we use the minimum value of f'(c)$: $f(5) \ge 7 + 4(-2) = 7 - 8 = -1$

2. Upper Bound: To find the maximum possible value for $f(5), we use the maximum value of f'(c)$: $f(5) \le 7 + 4(3) = 7 + 12 = 19$

Combining these results, we find that the range of possible values for $f(5)$ is $-1 \le f(5) \le 19$.

To evaluate the integral $\int_{1}^{e} \frac{(\ln x)^2}{x} dx$, we use the method of $u$-substitution.

1. Let $u = \ln x. Then, the derivative is du = \frac{1}{x} dx$.
2. Next, we change the limits of integration according to our substitution:
   • When $x = 1$, $u = \ln(1) = 0$.
   • When $x = e$, $u = \ln(e) = 1$.
3. Now, rewrite the integral in terms of $u$: $\int_{0}^{1} u^2 du$
4. Find the antiderivative of $u^2$, which is $\frac{u^3}{3}$.
5. Evaluate the antiderivative at the new limits: $\left[ \frac{u^3}{3} \right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}$

Answer: C

To evaluate the limit $\lim_{n \to \infty} \left(1 + \frac{3}{n}\right)^{2n}$, first identify the indeterminate form. As $n$ approaches infinity, $\frac{3}{n} approaches 0$, making the base approach 1, while the exponent 2n approaches infinity, $resulting in the form 1^{\infty}. We utilize the fundamental limit identity \lim_{n \to \infty} (1 + \frac{k}{n})^n = e^k. By applying exponent laws$, $the expression can be rewritten as a_n = \left[\left(1 + \frac{3}{n}\right)^n\right]^2. Applying the limit to the inner part gives \lim_{n \to \infty} a_n = [e^3]^2 = e^6. Alternatively$, $taking the natural logarithm yields \ln(a_n) = 2n \ln(1 + \frac{3}{n}). Using the limit comparison$ or L'Hôpital's Rule, $\lim_{n \to \infty} 2n \cdot \frac{3}{n} = 6$, so $a_n \to e^6. The final answer is **e^6$**.

To evaluate $\int x e^{5x} dx$, we use the Integration by Parts formula: $\int u \, dv = uv - \int v \, du$

1. **Identify $u$ and $dv**: Using the LIATE rule, we choose the algebraic term for u and the exponential term for dv$. Let $u = x \implies du = dx$ Let $dv = e^{5x} dx \implies v = \int e^{5x} dx = \frac{1}{5}e^{5x}$

2. Apply the formula: $\int x e^{5x} dx = (x)\left(\frac{1}{5}e^{5x}\right) - \int \left(\frac{1}{5}e^{5x}\right) dx$ $\int x e^{5x} dx = \frac{1}{5}x e^{5x} - \frac{1}{5} \int e^{5x} dx$

3. Evaluate the remaining integral: $\int x e^{5x} dx = \frac{1}{5}x e^{5x} - \frac{1}{5} \left(\frac{1}{5}e^{5x}\right) + C$ $\int x e^{5x} dx = \frac{1}{5}x e^{5x} - \frac{1}{25}e^{5x} + C$

The correct result is $\frac{1}{5}x e^{5x} - \frac{1}{25}e^{5x} + C$.

To evaluate the discontinuities, we examine the behavior of the function at the critical points $x=0$ and $x=1$.

1. **At $x = 0**: The left-hand limit is \lim_{x \to 0^-} \sin\left(\frac{1}{x}\right)$. As $x approaches 0 from the left, \frac{1}{x}$ approaches $-\infty, causing the sine function to oscillate infinitely between -1 and 1. Since the limit does not exist due to this behavior, x = 0$ is an oscillatory (essential) discontinuity.

2. **At $x = 1**: We evaluate the one-sided limits using the expression \frac{x^2 - x}{|x - 1|} + 2$.

• Left-hand limit ($x \to 1^-$): Since $x < 1$, $|x - 1| = -(x - 1)$. Thus, $\lim_{x \to 1^-} \left(\frac{x(x - 1)}{-(x - 1)} + 2\right) = \lim_{x \to 1^-} (-x + 2) = 1$.
• Right-hand limit ($x \to 1^+$): Since $x > 1$, $|x - 1| = x - 1$. Thus, $\lim_{x \to 1^+} \left(\frac{x(x - 1)}{x - 1} + 2\right) = \lim_{x \to 1^+} (x + 2) = 3$.
• Since both one-sided limits are finite but not equal ($1 \neq 3$$), x = 1$ is a jump discontinuity.

Therefore, $x = 0$ is oscillatory and $x = 1$ is a jump discontinuity. The correct answer is A.

To classify the discontinuities at $x = 1, we must evaluate the one-sided limits for each function. For f(x): 1. The left-hand limit is \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (\frac{-(x-1)}{x-1} + 2) = -1 + 2 = 1. 2. The right-hand limit is \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{(x+3)(x-1)}{x-1} = \lim_{x \to 1^+} (x+3) = 4. Since the one-sided limits \lim_{x \to 1^-} f(x) = 1$ and $\lim_{x \to 1^+} f(x) = 4$ are both finite but not equal, $f(x) has a **jump discontinuity**. For g(x): 1. The left-hand limit is \lim_{x \to 1^-} \frac{\sin(x-1)}{x-1}$. Let $u = x-1$; as $x \to 1^-$, $u \to 0^-. Using the standard limit \lim_{u \to 0} \frac{\sin(u)}{u} = 1, the left-hand limit is 1. 2. The right-hand limit is \lim_{x \to 1^+} \frac{1}{x-1}$. As $x approaches 1 from the right, the denominator is a small positive value, so \lim_{x \to 1^+} g(x) = +\infty. Since at least one of the one-sided limits is infinite, g(x) has an **infinite discontinuity**. Therefore, f(x) has a jump discontinuity and g(x)$ has an infinite discontinuity. The correct option is B.

To determine differentiability at $x=0, we must evaluate the limit of the difference quotient, as standard rules like the product rule require differentiability of individual factors (which |x|$ lacks at $x=0$).

The derivative is defined as: $f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h}$

Step 1: Substitute the function. We have $f(0) = 0|0| = 0$ and $f(h) = h|h|$. Substituting these into the limit: $f'(0) = \lim_{h \to 0} \frac{h|h| - 0}{h}$

Step 2: Simplify the expression. For $h \neq 0$, we can cancel $h$ in the numerator and denominator: $\frac{h|h|}{h} = |h|$

Step 3: Evaluate the limit. Now we evaluate the limit of the simplified expression: $f'(0) = \lim_{h \to 0} |h|$

As $h approaches 0 from either the positive or negative side, |h|$ approaches 0.

Since the limit exists and equals 0, $f(x)$ is differentiable at $x=0$ and $f'(0) = 0$.

To solve for $\lim_{x \to 0} f(x)$, we analyze the given inequality using the Squeeze Theorem.

1. Convert the Absolute Value Inequality: The expression $|f(x) - 5| \leq g(x)$ is equivalent to: $-g(x) \leq f(x) - 5 \leq g(x)$ Substituting $g(x) = x^2 \sin^2\left(\frac{1}{x}\right) + \sqrt{|x|}$, we get: $-\left(x^2 \sin^2\left(\frac{1}{x}\right) + \sqrt{|x|}\right) \leq f(x) - 5 \leq x^2 \sin^2\left(\frac{1}{x}\right) + \sqrt{|x|}$

2. Isolate $f(x)$: Add 5 to all parts of the inequality: $5 - \left(x^2 \sin^2\left(\frac{1}{x}\right) + \sqrt{|x|}\right) \leq f(x) \leq 5 + \left(x^2 \sin^2\left(\frac{1}{x}\right) + \sqrt{|x|}\right)$

3. Evaluate the Limit of the Bounding Functions: We need to find $\lim_{x \to 0} \left(x^2 \sin^2\left(\frac{1}{x}\right) + \sqrt{|x|}\right)$.

• Since $0 \leq \sin^2\left(\frac{1}{x}\right) \leq 1$$ for all $$x \neq 0$$, it follows that $0 \leq x^2 \sin^2\left(\frac{1}{x}\right) \leq x^2.
• By the Squeeze Theorem, because $\lim_{x \to 0} 0 = 0$ and $\lim_{x \to 0} x^2 = 0$, then $\lim_{x \to 0} x^2 \sin^2\left(\frac{1}{x}\right) = 0$.
• Additionally, $\lim_{x \to 0} \sqrt{|x|} = 0$. Therefore, the limit of the entire expression $g(x)$ is $0 + 0 = 0$.