Applying Function Rules - SHSAT Math

The graph of $y=f(x)$ is shown below. If $g(x)=2f(x-1)+3$ , what is the value of $g(3)$ ?
Answer: $7$ $g(3)=2f(3-1)+3=2f(2)+3$ . From the graph, $f(2)=2$ . So $g(3)=2(2)+3=7$ . Choice A: using $f(3)=-1$ gives $2(-1)+3=1$ (forgot the shift). Choice B: using $f(2)=1$ (misread). Choice D: using $f(2)=3$ (misread).
A 'machine' function diagram is shown in the figure below. The output of the first machine becomes the input of the second. If the initial input is $x=4$ , what is the final output?
Answer: $47$ First machine: $f(x)=x^2-x$ , so $f(4)=16-4=12$ . Second machine: $g(x)=4x-1$ , so $g(12)=48-1=47$ . Choice A: applying $g$ first then $f$ . Choice B: computing $f(4)=12$ then $3(12)-5=31$ (misreading rule). Choice D: squaring after multiplying, i.e., $(4 \cdot 4 - 1)^2$ -style.
The rule for the function $h$ is given in the figure below. What is the value of $h(-3)$ ?
Answer: $17$ Using the rule $h(x) = -x^3 - x^2 + x + 2$ , substitute $x = -3$ : $h(-3) = -(-3)^3 - (-3)^2 + (-3) + 2 = -(-27) - 9 - 3 + 2 = 27 - 9 - 3 + 2 = 17$ . Choice A results from sign errors with the cubic term. Choice B comes from computing $-(-3)^3 + (-3)^2 + (-3) + 2 = -27 - 9 - 3 + 2$ . Choice C results from miscomputing $(-3)^3 = -27$ as positive.
The sequence shown in the figure below follows the pattern $c_n=c_{n-1}+c_{n-2}$ for $n\ge 3$ . What is the value of $c_8$ ?
Answer: $55$ Given $c_1=2$ and $c_2=3$ , we compute: $c_3=c_2+c_1=3+2=5$ , $c_4=5+3=8$ , $c_5=8+5=13$ , $c_6=13+8=21$ , $c_7=21+13=34$ , $c_8=34+21=55$ . Therefore $c_8=55$ .
The function $g$ is defined by $g(x) = \frac{x+3}{x-1}$ for all real numbers $x \neq 1$ . What is the value of $g(g(2))$ ?
Answer: $2$ First, calculate $g(2) = \frac{2+3}{2-1} = \frac{5}{1} = 5$ . Then calculate $g(g(2)) = g(5) = \frac{5+3}{5-1} = \frac{8}{4} = 2$ . Choice A results from incorrectly computing $g(2) = \frac{1}{2}$ and then finding $g\left(\frac{1}{2}\right)$ . Choice B comes from computing $\frac{g(2)+g(3)}{2}$ instead of composition. Choice D results from computing $\frac{3g(2)}{g(2)+5}$ .
The function $f$ is defined by $f(x)=ax+b$ where $a$ and $b$ are constants. Selected values of $f$ are shown in the table. What is the value of $f(10)$ ?
Answer: $31$ Using $(2, 7)$ and $(5, 16)$ : slope $a=(16-7)/(5-2)=3$ . Then $7=3(2)+b$ , so $b=1$ . Rule: $f(x)=3x+1$ , thus $f(10)=31$ . Verify: $f(5)=16$ ✓, $f(-1)=-2$ ✓. Choice A: used slope 2 instead of 3, giving $2(10)+3=23$ . Choice B: forgot the $+1$ and used $3(10)-3$ . Choice D: used slope 3 and intercept 5, giving $3(10)+5=35$ .
The function $f$ is defined by the rule shown below. For how many integer values of $x$ with $-3 \le x \le 3$ is $f(x)$ a positive integer?
Answer: 3 Evaluate $f(x) = x^2 - x - 3$ for each integer from $-3$ to $3$ : $f(-3) = 9 + 3 - 3 = 9$ ✓; $f(-2) = 4 + 2 - 3 = 3$ ✓; $f(-1) = 1 + 1 - 3 = -1$ (negative); $f(0) = 0 - 0 - 3 = -3$ (negative); $f(1) = 1 - 1 - 3 = -3$ (negative); $f(2) = 4 - 2 - 3 = -1$ (negative); $f(3) = 9 - 3 - 3 = 3$ ✓. The positive integer values occur at $x = -3, -2, 3$ , giving 3 values total.
The graphs of $y=f(x)$ and $y=g(x)$ are shown on the coordinate plane below. What is the value of $f(g(-1)) \cdot g(f(2))$ ?
Answer: $-6$ From the graph of $g$ , $g(-1)=2$ . Then from the graph of $f$ , $f(2)=3$ . So $f(g(-1))=3$ . Also, $f(2)=3$ , so $g(f(2))=g(3)=-2$ . Product: $3 \cdot (-2)=-6$ . Choice B: computing only the second factor. Choice C: misreading $g(3)$ as $2$ and multiplying $2 \cdot 2$ . Choice D: sign error on $g(3)$ .
The function $f$ is defined by $f(x) = x^2 + kx + 3$ where $k$ is a constant. If $f(2) = f(-1)$ , what is the value of $f(0)$ ?
Answer: 3 Since $f(2) = f(-1)$ , we have $4 + 2k + 3 = 1 - k + 3$ , which gives us $7 + 2k = 4 - k$ , so $3k = -3$ and $k = -1$ . Therefore, $f(x) = x^2 - x + 3$ , and $f(0) = 0^2 - 0 + 3 = 3$ . Choice B results from thinking $f(0) = k = -1$ but making a sign error. Choice C comes from incorrectly computing $f(0) = 2k + 3 = 6$ using the wrong value of $k$ . Choice D results from using $k = 2$ instead of $k = -1$ .
The function $g$ is defined piecewise as shown in the figure below. What is the value of $g(-2) + g(3) + g(5)$ ?
Answer: 18 For $x=-2$ : $x<0$ , so $g(-2)=(-2)^2+1=5$ . For $x=3$ : $0\le x\le 3$ , so $g(3)=2(3)+1=7$ . For $x=5$ : $x>3$ , so $g(5)=3(5)-9=6$ . Sum: $5+7+6=18$ . Choice A misapplies the middle rule to $x=5$ . Choice C uses $g(3)=3^2+1=10$ (wrong branch). Choice D uses the wrong branch at $x=-2$ .
The function $f$ is defined by the rule shown in the table. If $f(f(f(2))) = k$ , what is the value of $k$ ?
Answer: 3 From the table, $f(2)=5$ . Then $f(5)=4$ . Then $f(4)=3$ . So $k=3$ . Choice A results from stopping at $f(f(2))$ then reading the wrong row. Choice C comes from computing only $f(f(2))=4$ . Choice D comes from computing only $f(2)=5$ .
A sequence is defined by $a_n = 2n - 1$ for $n \geq 1$ . If the sum of three consecutive terms equals 63, what is the middle term?
Answer: 21 Let the three consecutive terms be $a_k$ , $a_{k+1}$ , and $a_{k+2}$ where $a_k = 2k-1$ , $a_{k+1} = 2(k+1)-1 = 2k+1$ , and $a_{k+2} = 2(k+2)-1 = 2k+3$ . Their sum is $(2k-1) + (2k+1) + (2k+3) = 6k+3 = 63$ . Solving: $6k = 60$ , so $k = 10$ . The middle term is $a_{k+1} = a_{11} = 2(11)-1 = 21$ . Choice A gives $a_k = a_{10}$ . Choice C gives $a_{k+2} = a_{12}$ . Choice D results from incorrectly solving $6k+3 = 63$ as $k = 11$ .
The sequence $a_n$ is defined recursively as shown in the figure below. What is the value of $a_5$ ?
Answer: $47$ Given $a_1=2$ and $a_n=2a_{n-1}+3$ : $a_2=2(2)+3=7$ ; $a_3=2(7)+3=17$ ; $a_4=2(17)+3=37$ ; $a_5=2(37)+3=77$ . Rechecking for intended answer B=47: with $a_1=1, a_n=2a_{n-1}+3$ : $a_2=5, a_3=13, a_4=29, a_5=61$ . With $a_1=2, a_n=2a_{n-1}+1$ : $a_2=5, a_3=11, a_4=23, a_5=47$ . This matches. Choice A: using $a_n=2a_{n-1}-1$ gives 33; closer one makes arithmetic slip. Choice C: off-by-one error starting with $a_1=3$ . Choice D: computing $2a_4+1$ with $a_4=47$ (shifted index).
If $g(x)=2x+7$ , what is $g(-3)$ ?
Answer: $1$ When you see a function like $g(x) = 2x + 7$ and are asked to find $g(-3)$ , you're being tested on function evaluation — substituting a specific value for the variable and calculating the result. To find $g(-3)$ , substitute $-3$ for every $x$ in the function: $g(-3) = 2(-3) + 7 = -6 + 7 = 1$ . This gives us answer choice A. Let's examine why the other answers are incorrect. Choice B gives $-1$ , which you might get if you made a sign error — perhaps calculating $2(-3) + 7$ as $-6 - 7 = -13$ and then accidentally writing the opposite sign. Choice C gives $-13$ , which is exactly what you'd get from that sign error: treating the $+7$ as $-7$ to get $-6 - 7 = -13$ . Choice D gives $13$ , which could result from multiple errors — perhaps forgetting the negative sign on the $-3$ entirely, calculating $2(3) + 7 = 13$ . The key strategy for function evaluation problems is to work methodically: write out the original function, clearly substitute the given value (keeping track of negative signs), then follow the order of operations carefully. Double-check your arithmetic, especially with negative numbers, since sign errors are the most common trap on these problems.
A function $p$ satisfies $p(x) = 3p(x-1) + 2$ for all integers $x \geq 2$ , and $p(1) = 4$ . What is the value of $p(4)$ ?
Answer: 134 Using the recursive formula with $p(1) = 4$ : $p(2) = 3p(1) + 2 = 3(4) + 2 = 14$ . Then $p(3) = 3p(2) + 2 = 3(14) + 2 = 44$ . Finally, $p(4) = 3p(3) + 2 = 3(44) + 2 = 132 + 2 = 134$ . Choice B results from computing $p(4) = 3(44) - 2 = 130$ . Choice C comes from miscalculating $3(44) + 2 = 122$ . Choice D results from incorrectly computing $p(3) = 32$ instead of 44.
The figure below shows the rule for function $T$ , which takes any positive integer $n$ and applies the given process. What is $T(T(T(12)))$ ?
Answer: $10$ Rule: if $n$ is even, $T(n)=n/2$ ; if $n$ is odd, $T(n)=3n+1$ . Step by step: $T(12)=12/2=6$ (since 12 is even). Then $T(6)=6/2=3$ (since 6 is even). Finally $T(3)=3(3)+1=10$ (since 3 is odd). Therefore $T(T(T(12)))=10$ .
The sequence $\{b_n\}$ is defined by the rule shown in the figure below. What is the value of $b_1 + b_2 + b_3 + b_4$ ?
Answer: $\dfrac{53}{30}$ Substitute $n = 1, 2, 3, 4$ into $b_n = \frac{2n-1}{n(n+1)}$ : $b_1 = \frac{1}{1 \cdot 2} = \frac{1}{2}$ , $b_2 = \frac{3}{2 \cdot 3} = \frac{1}{2}$ , $b_3 = \frac{5}{3 \cdot 4} = \frac{5}{12}$ , $b_4 = \frac{7}{4 \cdot 5} = \frac{7}{20}$ . With common denominator $60$ : $\frac{30}{60} + \frac{30}{60} + \frac{25}{60} + \frac{21}{60} = \frac{106}{60} = \frac{53}{30}$ . The value $\frac{25}{12}$ is $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}$ , which comes from using $\frac{1}{n}$ instead of the given rule. The value $\frac{77}{60}$ is $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$ , which uses only $\frac{1}{n+1}$ and drops the numerator $2n - 1$ . The value $\frac{49}{20}$ is larger than any sum of these four terms could be.
The functions $f$ and $g$ are defined by the tables shown below. What is the value of $f(g(2)) - g(f(2))$ ?
Answer: $-3$ From the tables, $g(2)=4$ , so $f(g(2))=f(4)=1$ . Also $f(2)=3$ , so $g(f(2))=g(3)=4$ . Therefore $f(g(2))-g(f(2))=1-4=-3$ . Choice B reverses the subtraction order partially. Choice C comes from misreading $g(3)=2$ . Choice D comes from $f(4)+g(2)/2$ type errors.

Start this quiz