Free math worksheets

- Squares are never negative. - A negative sign outside a positive expression reverses its sign. - Determine whether the quantity being squared can equal \(0\).

1. a) Since \(x\) and \(y\) are nonzero, both squares are positive, so their sum is always positive. 2. b) Since \(x^2 + 3 \ge 3\), its opposite is always negative. 3. c) A square is nonnegative, and it equals \(0\) when \(x = y\), which is allowed. 4. d) Since \(x^4 \ge 0\), \(x^4 + 10\) is always positive. 5. e) Since \(x^2 + y^2 > 0\), its opposite is always negative.

a) Always positive b) Always negative c) Can equal zero when \(x = y\) d) Always positive e) Always negative

- Test positive values, negative values, and \(0\). - Compare the signs of even and odd powers. - The least possible value of a square is \(0\).

1. a) No. Since \(a^2 \ge 0\), \(a^2 + 5 \ge 5\). 2. b) No. A square is never negative. 3. c) Yes. For example, \((-1)^3 = -1\). 4. d) Yes. For example, \(0^2 - 10 = -10\). 5. e) Yes. Since \((a - 2)^2 \ge 0\), the expression is always at most \(-1\), so it is always negative.

a) No; \(a^2+5\ge5>0\). b) No; \((a+b)^2\ge0\). c) Yes; for example, \(a=-1\) gives \(a^3=-1\). d) Yes; for example, \(a=0\) gives \(a^2-10=-10\). e) Yes; \(-(a-2)^2-1\le-1\), so the expression is always negative.

- Determine the least possible value of \(x^2\). - Then add \(1\) to that lower bound.

1. For every real number \(x\), \(x^2 \ge 0\). 2. Therefore, \(x^2 + 1 \ge 1\). 3. Since \(1 > 0\), the inequality is true for every real \(x\).

Yes. Since \(x^2 \ge 0\), \(x^2 + 1 \ge 1 > 0\).

- Test several types of real numbers, including one between \(0\) and \(1\). - Squaring a positive number between \(0\) and \(1\) makes it smaller.

1. Test a real number between \(0\) and \(1\), such as \(a = 0.5\). 2. Then \(a^2 = 0.25\), and \(0.25 \ge 0.5\) is false. 3. Therefore, the inequality is not true for every real number.

No. For \(a = 0.5\), \(a^2 = 0.25 < 0.5\).

- What values are not allowed as the argument of a real logarithm? - Factor the quadratic expression to find its zeros. - Use the sign of the quadratic on the intervals determined by its zeros.

1. A real logarithm is undefined when its argument is less than or equal to \(0\), so solve \(x^2 - 7x + 10 \le 0\). 2. Factor the quadratic: \(x^2 - 7x + 10 = (x - 2)(x - 5)\). 3. The product is nonpositive between the zeros, including the zeros. Therefore, \(2 \le x \le 5\).

\(x \in [2, 5]\)

- Evaluate both functions at an input between \(0\) and \(1\). - Compare higher powers when the base is greater than \(1\). - Use the point \((1, 1)\) as the boundary between the two behaviors.

1. \(f(0.5)=(0.5)^2=0.25\), while \(g(0.5)=(0.5)^4=0.0625\). 2. Graph \(u\) is higher at \(x=0.5\), so graph \(u\) represents \(f(x)=x^2\), and graph \(v\) represents \(g(x)=x^4\). 3. For \(x>1\), \(x^4>x^2\), so graph \(v\) lies above graph \(u\).

a) Graph \(u\): \(f(x)=x^2\); graph \(v\): \(g(x)=x^4\) b) For \(x>1\), graph \(v\) lies above graph \(u\) because \(x^4>x^2\).

- Set the two function expressions equal to find their intersection points. - Factor the resulting polynomial equation. - To compare the graphs, solve \(f(x)<g(x)\). - Test intervals separated by the values where the functions are equal.

1. Set the functions equal: \(\frac{1}{2}x^4=8x^2\). 2. Move all terms to one side and factor: \(\frac{1}{2}x^4-8x^2=0\), so \(x^2\left(\frac{1}{2}x^2-8\right)=0\). 3. The solutions are \(x=0\) or \(x^2=16\), so \(x=-4, 0, 4\). 4. The corresponding y-values are \(0\) and \(128\). The intersection points are \((-4, 128)\), \((0, 0)\), and \((4, 128)\). 5. Solve \(\frac{1}{2}x^4<8x^2\). Factoring gives \(\frac{1}{2}x^2(x^2-16)<0\). Because \(x^2>0\) for \(x\ne0\), this occurs when \(-4<x<4\), excluding \(x=0\), where the functions are equal.

The intersection points are \((-4, 128)\), \((0, 0)\), and \((4, 128)\). The graph of \(f\) lies below the graph of \(g\) on \((-4, 0) \cup (0, 4)\).

- Remember that \(x^2\) is nonnegative for every real \(x\). - What sign does \(x^3\) have when \(x<0\)? - Test the intervals to the left of \(0\), between \(0\) and \(4\), and to the right of \(4\).

1. Set the functions equal: \(x^3=4x^2\). 2. Factor: \(x^2(x-4)=0\), so \(x=0\) or \(x=4\). 3. The corresponding points are \((0, 0)\) and \((4, 64)\). 4. Solve \(4x^2>x^3\): \(x^2(4-x)>0\). 5. Since \(x^2>0\) for \(x\ne0\), the inequality holds when \(4-x>0\), excluding \(x=0\). Therefore, \(f(x)>g(x)\) on \((-\infty, 0)\cup(0, 4)\).

The intersection points are \((0, 0)\) and \((4, 64)\). The graph of \(f\) lies above the graph of \(g\) on \((-\infty, 0) \cup (0, 4)\).

- Move all terms to one side and factor. - Compare values with absolute value greater than \(1\) to values between \(-1\) and \(1\). - What signs can even powers have? - Can a fourth power be negative?

1. \(x^4-x^2=x^2(x^2-1)=x^2(x-1)(x+1)=0\), so \(x=-1,0,1\). 2. Solve \(x^2(x^2-1)>0\). Since \(x^2\ge0\), the expression is positive when \(x^2>1\), so \(x<-1\) or \(x>1\). 3. Solve \(x^2(x^2-1)<0\). This requires \(x\ne0\) and \(x^2<1\), so \(-1<x<0\) or \(0<x<1\). 4. Because \((-x)^2=x^2\) for every real \(x\), all real numbers satisfy the equation. 5. A fourth power is always nonnegative, so there is no real solution.

1) \(\{-1,0,1\}\) 2) \((-\infty, -1) \cup (1, \infty)\) 3) \((-1, 0) \cup (0, 1)\) 4) All real numbers 5) No real solution

- Factor \(x^5-x^3\) to find the critical values. - Compare powers when the base is greater than \(1\) and when it is between \(0\) and \(1\). - For negative bases, compare both the sign and absolute value. - Test values such as \(2\), \(0.5\), and \(-2\).

1. \(x^5-x^3=x^3(x^2-1)=x^3(x-1)(x+1)=0\), so \(x=-1,0,1\). 2. A sign analysis of \(x^3(x-1)(x+1)\) shows that it is positive on \((-1, 0)\) and \((1, \infty)\). 3. For \(x>1\), multiplying by the additional positive factor \(x^2>1\) makes \(x^5>x^3\). 4. For \(0<x<1\), multiplying by \(x^2<1\) makes \(x^5<x^3\). 5. For \(x<-1\), both powers are negative and \(|x^5|>|x^3|\), so \(x^5<x^3\).

1) \(\{-1,0,1\}\) 2) \((-1, 0)\cup(1, \infty)\) 3) a) \(x^5>x^3\) b) \(x^3>x^5\) c) \(x^3>x^5\)

- What condition must the logarithm's argument satisfy for every real input? - When is an upward-opening quadratic strictly above the x-axis? - Use the discriminant to determine when a quadratic has no real zeros. - Remember that an argument equal to \(0\) is not allowed.

1. The logarithm is defined for every real \(x\) only if \(x^2 + 6x + c > 0\) for every real \(x\). 2. The quadratic opens upward. It is always positive exactly when it has no real zeros, so its discriminant must be negative. 3. The discriminant is \(6^2 - 4(1)(c) = 36 - 4c\). Solve \(36 - 4c < 0\) to obtain \(c > 9\). 4. When \(c = 9\), the quadratic equals \((x + 3)^2\) and is \(0\) at \(x = -3\), so the inequality must be strict.

\(c > 9\)

- Find the x-intercepts by solving \(f(x)=0\). - Factor after clearing the decimal coefficient. - Use the sign of the leading coefficient to determine whether the parabola opens upward or downward. - The zeros divide the number line into intervals for a sign analysis.

1. Solve \(0.5x^2+2x-6=0\). Multiplying by \(2\) gives \(x^2+4x-12=0\), which factors as \((x+6)(x-2)=0\). The zeros are \(x=-6\) and \(x=2\). 2. The leading coefficient is positive, so the parabola opens upward. Therefore, the function is negative between the zeros and positive outside them. Thus, \(f(x)>0\) on \((-\infty, -6)\cup(2, \infty)\), and \(f(x)<0\) on \((-6, 2)\).

1. \(x=-6\) and \(x=2\) 2. \(f(x)>0\) on \((-\infty, -6)\cup(2, \infty)\); \(f(x)<0\) on \((-6, 2)\).

- Solve for the squared expression without expanding the vertex form. - The negative coefficient means the parabola opens downward. - Decide whether the graph lies above the x-axis between or outside the zeros.

1. Set \(g(x)=0\): \(-3(x-1)^2+27=0\). Then \((x-1)^2=9\), so \(x-1=\pm3\). The zeros are \(x=-2\) and \(x=4\). 2. The leading coefficient is negative, so the parabola opens downward. Its values are positive between the two zeros. Therefore, \(g(x)>0\) for \(-2<x<4\).

1. \(x=-2\) and \(x=4\) 2. \(g(x)>0\) on \((-2, 4)\).

- Find one integer zero, then divide to factor the polynomial completely. - The leading term determines the end behavior. - A polynomial changes sign when it passes through a zero of odd multiplicity. - Use the factored form or a test point in each interval to determine the sign.

1. Testing integer values gives the zero \(x = 1\). Dividing by \(x - 1\) gives \(x^2 - 3x - 10\), which factors as \((x - 5)(x + 2)\). Thus, \(f(x) = (x + 2)(x - 1)(x - 5)\), and the zeros are \(x = -2\), \(x = 1\), and \(x = 5\). 2. The leading term is \(x^3\), which has odd degree and a positive coefficient. Therefore, \(f(x) \to \infty\) as \(x \to \infty\), and \(f(x) \to -\infty\) as \(x \to -\infty\). 3. The simple zeros divide the number line into \((-\infty, -2)\), \((-2, 1)\), \((1, 5)\), and \((5, \infty)\). The sign changes at each simple zero. Testing one point in each interval gives \(f(x) < 0\) on \((-\infty, -2) \cup (1, 5)\) and \(f(x) > 0\) on \((-2, 1) \cup (5, \infty)\).

1. \(x = -2\), \(x = 1\), \(x = 5\) 2. \(f(x) \to \infty\) as \(x \to \infty\), and \(f(x) \to -\infty\) as \(x \to -\infty\). 3. \(f(x) > 0\) on \((-2, 1) \cup (5, \infty)\); \(f(x) < 0\) on \((-\infty, -2) \cup (1, 5)\).

- Look for a substitution that turns the fourth-degree equation into a quadratic equation. - Use the degree and sign of the leading coefficient to determine the end behavior. - Factor the polynomial completely after finding its zeros. - Test one value in each interval between consecutive zeros, or use the multiplicity of each zero to track sign changes.

1. Set \(g(x) = 0\) and let \(u = x^2\). Then \(-u^2 + 10u - 9 = 0\), or \(u^2 - 10u + 9 = 0\). 2. Factor: \((u - 1)(u - 9) = 0\), so \(u = 1\) or \(u = 9\). Back-substitution gives \(x = \pm 1\) and \(x = \pm 3\). 3. The leading term is \(-x^4\). Because its degree is even and its coefficient is negative, \(g(x) \to -\infty\) as \(x \to \infty\) and as \(x \to -\infty\). 4. The factorization is \(g(x) = -(x - 3)(x - 1)(x + 1)(x + 3)\). Each zero is simple, so the sign changes at every zero. 5. Therefore, \(g(x) > 0\) on \((-3, -1) \cup (1, 3)\), and \(g(x) < 0\) on \((-\infty, -3) \cup (-1, 1) \cup (3, \infty)\).

1. \(x \in \{-3, -1, 1, 3\}\) 2. \(g(x) \to -\infty\) as \(x \to \pm\infty\). 3. Positive on \((-3, -1) \cup (1, 3)\); negative on \((-\infty, -3) \cup (-1, 1) \cup (3, \infty)\).

- Substitute a parameter value to match the given function. - Compare the graph with the line by factoring \(f_k(x)-x\). - Use the multiplicity and sign of each factor to build a sign chart. - Treat the cases \(k>0\) and \(k<0\) separately.

1. Set \(k=2\): \(f_2(x)=\frac12x^3-2x^2+x=f(x)\). 2. Intersections with \(y=x\) satisfy \(f_k(x)-x=0\). Factor the difference: \(f_k(x)-x=\frac{k}{4}x^2(x-4)\). Since \(k\ne0\), the intersection x-coordinates are \(0\) and \(4\). The intersection points are \((0, 0)\) and \((4, 4)\). The root at \(x=0\) has multiplicity \(2\), so the graph is tangent to the line there. 3. The factor \(x^2\) is positive for \(x\ne0\), so the sign of \(f_k(x)-x\) is determined by \(k(x-4)\). For \(k>0\), the graph is below the line on \((-\infty, 0)\cup(0, 4)\) and above it on \((4, \infty)\). For \(k<0\), the graph is above the line on \((-\infty, 0)\cup(0, 4)\) and below it on \((4, \infty)\).

a) \(k=2\) b) The intersection points are \((0, 0)\) and \((4, 4)\). For \(k>0\): below on \((-\infty, 0)\cup(0, 4)\), above on \((4, \infty)\). For \(k<0\): above on \((-\infty, 0)\cup(0, 4)\), below on \((4, \infty)\).

- Set the two function expressions equal and factor. - Test an input such as \(x=0.5\). - Compare powers of numbers whose absolute value is less than \(1\).

1. Set the functions equal: \(x^3=x^5\), so \(x^3(1-x^2)=0\). Thus, \(x=-1,0,1\), giving the points \((-1, -1)\), \((0, 0)\), and \((1, 1)\). 2. For \(0<x<1\), \(x^5<x^3\). Therefore, the graph of \(g\) lies closer to the x-axis. 3. As the positive odd exponent increases, the graph becomes flatter and stays closer to the x-axis on \((-1, 1)\).

a) \((-1, -1)\), \((0, 0)\), and \((1, 1)\) b) The graph of \(g(x)=x^5\), because \(x^5<x^3\) for \(0<x<1\) c) The graph becomes flatter and closer to the x-axis on \((-1, 1)\).

- Set the two function expressions equal to find intersection points. - Compare powers of a number greater than \(1\) with powers of a number between \(0\) and \(1\). - Restrict the final answer to the given interval.

1. Set the functions equal: \(x^4=x^2\). 2. Factor: \(x^4-x^2=x^2(x^2-1)=0\). In \([0, 1.5]\), the solutions are \(x=0\) and \(x=1\), giving the points \((0, 0)\) and \((1, 1)\). 3. For \(x>1\), \(x^2>1\). Multiplying by \(x^2>0\) gives \(x^4>x^2\). 4. Therefore, within the stated interval, \(g(x)>f(x)\) for \(1<x\le1.5\).

a) \((0, 0)\) and \((1, 1)\) b) \((1, 1.5]\)

- Set the two function expressions equal to find intersections. - To compare the graphs, solve an inequality involving \(x^4-x^2\). - What conditions on \(a\) and \(n\) make the range consist of all nonnegative real numbers? - For part c), try different positive even values of \(n\).

1. Set the functions equal: \(x^2=x^4\), so \(x^4-x^2=0\). 2. Factor: \(x^2(x^2-1)=x^2(x-1)(x+1)=0\). Thus, \(x=-1, 0, 1\), giving the intersection points \((-1, 1)\), \((0, 0)\), and \((1, 1)\). 3. Solve \(x^4>x^2\): \(x^2(x^2-1)>0\). Since \(x^2\ge 0\), the inequality holds when \(x^2>1\), or \(x<-1\) and \(x>1\). 4. The range of \(f(x)=x^2\) is \([0,\infty)\). For \(h(x)=ax^n\) to have this range, \(n\) must be even and \(a>0\). 5. Using \(h(2)=16\), one choice is \(n=2\), which gives \(a=4\), and another is \(n=4\), which gives \(a=1\).

a) \((-1, 1)\), \((0, 0)\), and \((1, 1)\) b) \((-\infty, -1) \cup (1, \infty)\) c) Two possible rules are \(h(x)=4x^2\) and \(h(x)=x^4\).

- Rewrite both sides as powers with base \(2\). - Express \(0.5\) as the reciprocal of \(2\). - Because the common base is greater than \(1\), the inequality direction stays the same when comparing exponents. - Solve the resulting quadratic inequality by locating the zeros and testing intervals.

1. Rewrite both sides with base \(2\): \((2^{-1})^{x^2-1}\le(2^2)^{x-1}\). 2. Simplify the exponents: \(2^{-x^2+1}\le2^{2x-2}\). 3. Since the base \(2>1\), compare the exponents without reversing the inequality: \(-x^2+1\le2x-2\). 4. Rearrange: \(x^2+2x-3\ge0\). 5. Factor: \((x+3)(x-1)\ge0\). The product is nonnegative outside the interval between its zeros. 6. Therefore, \(x\le-3\) or \(x\ge1\).

\((-\infty, -3] \cup [1, \infty)\)