Exercises: Prove Growth Properties of Functions

$5x + 3 + 2 = 5x + 5$

$5(x + 3) + 2 = 5x + 17$

$5x + 2 + 3 = 5x + 5$

$(5x + 2) \cdot 3 = 15x + 6$

$b^{x+d} = b^x + b^d$

$b^{x+d} = b^x \cdot b^d$

$b^{x+d} = (b+b)^{x+d} = (2b)^{x+d}$

$b^{x+d} = b^{xd}$

The starting value when $x = 0$.

The constant rate of change — how much $f$ increases per unit increase in $x$.

The $y$-value at the maximum.

The ratio of consecutive outputs.

$-3(x + 2) + 10 - (-3x + 10) = -6$

$-3x + 2 + 10 - (-3x + 10) = 2$

$(-3x + 10) \cdot 2 = -6x + 20$

$-3(x + 2) - (-3x) = -6x - 6$

$\frac{5 \cdot 3^{x+2}}{5 \cdot 3^x} = \frac{3^x \cdot 3^2}{3^x} = 9$

$\frac{5 \cdot 3^x + 5 \cdot 3^2}{5 \cdot 3^x} = 1 + \frac{9}{3^x}$, which depends on $x$

$\frac{5 \cdot 3^{2x}}{5 \cdot 3^x} = 3^x$, which depends on $x$

$\frac{5 \cdot 3^{x+2}}{5 \cdot 3^x} = x + 2$

The slope $m$ equals the constant difference divided by the interval width $d$.

The slope $m$ equals the constant difference (the result $md$ itself).

The slope $m$ is irrelevant to the constant difference.

The slope $m$ equals the interval width $d$.

Equal differences (each output increases by 6), suggesting a linear function.

Equal ratios (each output is multiplied by 2), suggesting an exponential function.

Equal differences (each output increases by 6) AND equal ratios (each output doubles), suggesting both linear and exponential.

Neither equal differences nor equal ratios — this is a quadratic function.

Step 3 is wrong — the result should be $7d$, not $d$.

Step 1 is wrong — $f(x + d) = 7(x + d) - 3 = 7x + 7d - 3$, not $7x + d - 3$.

Step 2 is wrong — you cannot subtract $f(x)$ from $f(x + d)$ in a proof.

The proof is correct. The constant difference is $d$.

The student made an algebra error — the correct difference is $4 \cdot (1.5)^2 = 9$, which is constant.

The student used the wrong test. Exponential functions have constant RATIOS, not constant differences. The student should compute $g(x+2) / g(x)$.

The function is not actually exponential, so neither the difference nor the ratio will be constant.

The student should have used $d = 1$ instead of $d = 2$.

Yes — a numerical example that confirms the formula is sufficient as a proof.

No — this is verification of one case. A proof must use a VARIABLE $x$ and show the result holds for ALL $x$.

No — the student needs to check more cases before it counts as a proof.

Yes — if the formula works for $x = 2$, it will work for all non-negative integers.

The student computed $f(13)$ incorrectly.

The student only verified one specific case ($x = 10$, $d = 3$). A proof must use a variable $x$ to show the result holds for all $x$.

The student should have computed $f(x + d) / f(x)$, not $f(x + d) - f(x)$.

The proof is valid — checking one case is sufficient for linear functions.

Step 3 — the conclusion is wrong. The function IS exponential.

Step 1 — the exponent rule gives $3^{x+2} = 3^x \cdot 3^2$, not $3^x + 3^2$.

Step 2 — you cannot divide $g(x+2)$ by $g(x)$.

There is no error — the function is not exponential.

Yes — since $5 \cdot 2^x$ is exponential and we just added a constant, the function is still exponential.

Yes — the base is 2, so the ratio is always 2.

No — compute $p(x+1)/p(x) = (10 \cdot 2^x + 3)/(5 \cdot 2^x + 3)$, which depends on $x$ and is not constant.

No — adding a constant to any function makes it non-exponential, and exponential functions can never have constant ratios.