Solving Nonlinear Functions - SAT Math

A function is defined by $g(x)=\sqrt{5-x}$ . For real outputs, which interval gives the domain of $g$ ?
Answer: $(-\infty,5]$ To find the domain of g(x) = √(5-x), we need to determine when the expression under the square root is non-negative. For real outputs, we need 5 - x ≥ 0, which means 5 ≥ x or x ≤ 5. This gives us the domain (-∞, 5], which includes all real numbers less than or equal to 5. The common mistake is reversing the inequality when solving for x or forgetting that the square root requires non-negative values. When working with square root functions, always set the expression under the radical greater than or equal to zero.
Let $h(x)=\dfrac{x+3}{x-2}$ . Which value is not in the domain of $h$ , and what is $h(5)$ ?
Answer: $x\ne 2$ ; $h(5)=\tfrac{8}{3}$ The question asks for the value not in the domain of h(x)=(x+3)/(x-2) and the value of h(5). This is a rational function, characterized by restrictions where the denominator is zero, unlike polynomials or exponentials with full domains. The domain excludes x=2, as it makes the denominator zero, causing undefined behavior. Now compute h(5)=(5+3)/(5-2)=8/3. Common errors include mistaking x=-3 as excluded (it makes numerator zero but is defined as 0) or miscalculating the fraction. Another error is confusing domain with range in rational functions. As a test-taking strategy, set the denominator to zero to find exclusions and plug in values directly for evaluation.
A function is defined by $r(x)=\sqrt{x-4}$ . What is the domain of $r(x)$ in real numbers?
Answer: $[4,\infty)$ The question requires the domain of r(x) = √(x - 4) in real numbers. This is a radical function, nonlinear and defined only for non-negative radicands in reals. For the square root to be real, x - 4 ≥ 0, so x ≥ 4, giving domain [4, ∞). This includes x=4 where r(4)=0. Errors include thinking it's all reals like linear, or excluding 4, but it's defined there. Another mistake is reversing inequality to x ≤ 4. For radicals, solve the inequality inside ≥ 0 and consider even roots restrict to non-negative.
A bacteria culture is modeled by the function $P(t)=120\cdot(1.25)^t$ , where $t$ is the number of hours after the start. Which statement best describes the growth shown by this model, and what is the growth factor per hour?
Answer: Exponential; multiplies by $1.25$ This question asks us to identify the function type and growth factor for P(t) = 120· $(1.25)^t$ . The function has the form a· $b^t$ where a = 120 and b = 1.25, which is the standard form of an exponential function. Since the base 1.25 > 1, this represents exponential growth where the population multiplies by 1.25 each hour. A common error is confusing the growth factor (1.25) with the growth rate (0.25 or 25%). When analyzing exponential models, look for the form a· $b^t$ and identify b as the multiplicative factor per time unit.
If $r(x)=\sqrt{9-2x}$ is real-valued, what is the largest possible value of $x$ in its domain?
Answer: $\dfrac{9}{2}$ We need to find the largest x-value in the domain of r(x) = √(9-2x). This is a square root function requiring 9 - 2x ≥ 0 for real values. Solving the inequality: 9 - 2x ≥ 0; 9 ≥ 2x; 9/2 ≥ x; x ≤ 9/2. The largest possible x-value is 9/2. We can verify: when x = 9/2, we get 9 - 2(9/2) = 9 - 9 = 0, and √0 = 0 is defined. Common errors include solving the inequality incorrectly or forgetting to flip the inequality sign when dividing by negative numbers (though we divided by positive 2 here).
A graph is shown. It starts at the point $(2,0)$ and increases slowly to the right, with no points for $x<2$ . Which equation best matches the graph?
Answer: $y=\sqrt{x-2}$ This question describes a graph starting at $(2, 0)$ and increasing slowly to the right with no points for $x < 2$ . This matches a square root function with a horizontal shift. Option A, $y = \sqrt{x-2}$ , starts at $x = 2$ (where $x - 2 = 0$ ) and increases slowly as $x$ increases. The function is undefined for $x < 2$ since we'd have the square root of a negative number. Options B (rational) would have a vertical asymptote, C (absolute value) would have a V-shape, and D (exponential) wouldn't start at exactly $(2, 0)$ . Radical functions have a characteristic starting point and gradual increase.
A function is defined by $r(x)=\dfrac{x+3}{x^2-9}$ . What is the domain of $r$ ? Choose the option that excludes values that make the denominator zero (even if factors cancel algebraically).
Answer: $(-\infty,-3)\cup(-3,3)\cup(3,\infty)$ This question asks for the domain of r(x) = (x + 3)/(x² - 9). Rational functions are undefined where the denominator equals zero. Setting x² - 9 = 0, we get x² = 9, so x = ±3. We can factor the denominator as (x - 3)(x + 3), confirming zeros at x = 3 and x = -3. Even though the numerator (x + 3) shares a factor with the denominator, we still exclude x = -3 from the domain because the original function is undefined there. The domain is (-∞, -3) ∪ (-3, 3) ∪ (3, ∞). A common error is canceling factors before determining domain restrictions. Always find domain restrictions from the original denominator before any simplification.
A function is defined by $f(x)=\sqrt{2x-5}$ . What is the domain of $f$ in real numbers?
Answer: $\left[\tfrac{5}{2},\infty\right)$ The question asks for the domain of the function $f(x)=\sqrt{2x-5}$ in real numbers. This is a square root (radical) function, distinguished by its requirement that the radicand be non-negative, unlike unrestricted domains in polynomials or exponentials. To find the domain, solve the inequality $2x-5 \geq 0$ , which gives $2x \geq 5$ , so $x \geq \frac{5}{2}$ . Thus, the domain is all real numbers from $\frac{5}{2}$ to infinity, including $\frac{5}{2}$ where $f(\frac{5}{2})=0$ . A common error is mistakenly allowing $x < \frac{5}{2}$ , leading to imaginary values, or excluding the boundary point. Another error confuses it with ranges of other functions like rationals. As a test-taking strategy, for radical functions, always set the inside expression $\geq 0$ and solve for x to determine the domain.
A coordinate plane shows the graph of a function passing through the points $(-2,4)$ , $(0,0)$ , and $(2,4)$ , with a sharp corner at $(0,0)$ and symmetry about the $y$ -axis. Which equation best matches the graph?
Answer: $y=|2x|$ We need to match the described graph to an equation, noting the sharp corner at (0,0) and symmetry about the y-axis. The function passes through (-2,4), (0,0), and (2,4), showing that when |x| = 2, y = 4. Testing y = |2x|: when x = -2, y = |2(-2)| = |-4| = 4; when x = 0, y = |0| = 0; when x = 2, y = |2(2)| = 4. The sharp corner at the origin and the V-shape with symmetry about the y-axis are characteristic features of absolute value functions. Students often confuse this with a parabola, but parabolas have smooth curves, not sharp corners.
A coordinate plane shows a parabola opening upward with vertex at $(1,-4)$ and passing through $(0,-3)$ . Which equation matches the graph?
Answer: $y=(x-1)^2-4$ We need to match an equation to a parabola with vertex at (1,-4) passing through (0,-3). This is a quadratic function in vertex form: y = a(x-h)² + k where (h,k) = (1,-4). So y = a(x-1)² - 4. To find a, use the point (0,-3): -3 = a(0-1)² - 4; -3 = a(1) - 4; -3 = a - 4; a = 1. Therefore, y = (x-1)² - 4. We can verify: vertex is at (1,-4) ✓, and when x = 0: y = (0-1)² - 4 = 1 - 4 = -3 ✓. The parabola opens upward since a = 1 > 0. When given vertex and another point, use vertex form to find the equation.
A bacteria culture has mass $m(t)=120(1.35)^t$ , where $t$ is time in hours. Which statement best describes the function and its behavior over time?
Answer: Exponential; growth This question asks us to identify the type of function and describe its behavior for a bacteria culture with mass m(t) = 120 $(1.35)^t$ . The function has the form f(t) = a· $b^t$ where a = 120 and b = 1.35, which is the standard form of an exponential function. Since the base b = 1.35 > 1, this represents exponential growth - the bacteria mass increases by a factor of 1.35 each hour. The mass starts at 120 grams when t = 0 and grows without bound as time increases. A common error is confusing this with linear growth, but linear functions add a constant amount while exponential functions multiply by a constant factor. When you see a variable in the exponent with a base greater than 1, recognize it as exponential growth.
A function is defined by $r(x)=\dfrac{x+2}{x-4}$ . For what value of $x$ is $r(x)$ undefined?
Answer: $x=4$ This question asks where a rational function is undefined. The function r(x) = (x+2)/(x-4) is undefined when the denominator equals zero. Setting x - 4 = 0, we get x = 4. At this value, we would have division by zero, making the function undefined. The numerator's zeros (x = -2) determine where the function equals zero, not where it's undefined. A common error is setting the numerator equal to zero instead of the denominator. For rational functions, always find undefined points by setting the denominator equal to zero.
A coordinate plane shows a curve that crosses the $y$ -axis at $(0,3)$ and passes through $(1,6)$ and $(2,12)$ . Which equation best represents the curve?
Answer: $y=3\cdot 2^x$ We need to find an equation that passes through (0,3), (1,6), and (2,12). Notice that the y-values form the pattern 3, 6, 12, where each value doubles from the previous one - this suggests exponential growth. Testing y = 3· $2^x$ : when x = 0, y = 3· $2^0$ = 3·1 = 3; when x = 1, y = 3· $2^1$ = 6; when x = 2, y = 3· $2^2$ = 3·4 = 12. All three points match perfectly. The doubling pattern is the key indicator of an exponential function with base 2. When points show a constant multiplicative relationship, think exponential rather than linear or other function types.
A function is defined by $k(x)=\sqrt{x-4}-3$ . What is the value of $k(13)$ ?
Answer: $0$ This question asks us to evaluate k(13) where $k(x) = \sqrt{x - 4} - 3$ . Following the order of operations, we first work inside the radical: $k(13) = \sqrt{13 - 4} - 3 = \sqrt{9} - 3$ . Since $\sqrt{9} = 3$ , we get $k(13) = 3 - 3 = 0$ . The warning reminds us about the correct order: subtract 4 inside the radical first, then subtract 3 outside. A common error would be to subtract both values from x before taking the square root, or to subtract in the wrong order. When evaluating functions with multiple operations, work from the inside out: parentheses/radicals first, then outside operations.
A savings account balance is modeled by $B(t)=1500(1.04)^t$ , where $t$ is the number of years. Approximately what is the percent increase in the balance each year?
Answer: $4\%$ This question asks to interpret the growth rate in an exponential model. The function B(t) = 1500 $(1.04)^t$ has a growth factor of 1.04. Since 1.04 = 1 + 0.04, this represents a 4% increase each year. The initial balance is 1500, and each year it's multiplied by 1.04, growing by 4%. Common errors include interpreting 1.04 as 104% growth (which would more than double the balance) or as 0.04% growth. When the base of an exponential function is (1 + r), the growth rate is r expressed as a percentage.
A coordinate plane shows the graph of $y=\sqrt{x-1}$ . What is the domain of this function as shown (real outputs only)?
Answer: $[1,\infty)$ This question asks for the domain of y = √(x-1) when considering only real outputs. This is a square root function, which requires the expression under the radical to be non-negative. Setting up the inequality: x - 1 ≥ 0, which gives x ≥ 1. The domain is all x-values greater than or equal to 1, written as [1,∞). The square bracket indicates that x = 1 is included since √(1-1) = √0 = 0 is defined. A common error is using parentheses instead of brackets, which would exclude the endpoint. For radical functions with even roots, always use ≥ (not >) for the domain restriction.
On the coordinate plane, a curve passes through $(0,3)$ , decreases as $x$ increases, and approaches (but never reaches) the horizontal line $y=0$ . It also passes through $(1,1.5)$ . Which equation could represent the curve?
Answer: $y=3(0.5)^x$ The question requires selecting an equation for a curve passing through (0,3) and (1,1.5), decreasing toward y=0 without reaching it. This is an exponential decay function, distinguished by its asymptotic approach to the x-axis. Plug in x=0 to confirm y=3 matches the initial value, then x=1 gives y=3* $(0.5)^1$ =1.5, fitting the point and decay behavior. The curve approaches y=0 as x increases, consistent with base between 0 and 1. Errors include choosing growth like $(1.5)^x$ , which increases, or linear like 3-0.5x, which crosses y=0. Rational like 3/x passes (1,3) not (1,1.5). For such problems, test points in choices to verify fit and asymptotic behavior.
The function $f(x)=\left|2x-7\right|$ gives the distance between $2x$ and $7$ on the number line. What is the value of $f(1)$ ?
Answer: $5$ We need to find $f(1)$ where $f(x) = |2x - 7|$ . First, substitute $x = 1$ into the expression inside the absolute value bars: $2(1) - 7 = 2 - 7 = -5$ . Then apply the absolute value function: $|-5| = 5$ , so $f(1) = 5$ . The key error to avoid is forgetting to apply the absolute value after evaluating the expression inside. Remember that absolute value always returns a non-negative result, converting negative values to positive.
A medicine amount remaining is modeled by $A(t)=80(0.75)^t$ , where $t$ is in days. Approximately how much medicine remains after $2$ days?
Answer: $45$ We need to calculate A(2) where A(t) = 80 $(0.75)^t$ represents medicine amount after t days. Substituting t = 2: A(2) = 80 $(0.75)^2$ = 80(0.5625) = 45. Since 0.75 < 1, this is an exponential decay function where the medicine amount decreases by 25% each day (retaining 75%). After 2 days, only 56.25% of the original 80 units remain. The key to solving exponential problems is careful calculation of the base raised to the given power. Remember that $(0.75)^2$ means 0.75 × 0.75, not 0.75 × 2.
A function is defined by $g(x)=\sqrt{5-x}$ . For real outputs, which interval gives the domain of $g$ ?
Answer: $(-\infty,5]$ To find the domain of $g(x) = \sqrt{5-x}$ , we need to determine when the expression under the square root is non-negative. For real outputs, we need $5 - x \geq 0$ , which means $5 \geq x$ or $x \leq 5$ . This gives us the domain $(-\infty, 5]$ , which includes all real numbers less than or equal to 5. The common mistake is reversing the inequality when solving for x or forgetting that the square root requires non-negative values. When working with square root functions, always set the expression under the radical greater than or equal to zero.

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