{ "slug": "exponential-functions-and-graphs", "coreConcept": "An exponential function is of the form $f(x) = a^x$, where the **base** $a$ is a positive constant ($a > 0, a \\neq 1$) and the variable $x$ is the exponent. These functions are fundamental for modelling scenarios of rapid growth or decay and have distinct graphical properties based on the value of $a$.", "mainText": "### Introduction to Exponential Functions\n\nThe exponential function, written as $y = a^x$, is a cornerstone of advanced mathematics. In this form, $a$ is referred to as the **base** and is required to be a positive constant not equal to 1, while the variable $x$ can be any real number. The behaviour of the exponential function is critically dependent on the value of the base $a$, which determines whether the function represents growth or decay.\n\n### The Graph of $y = a^x$ for $a > 1$\n\nWhen the base $a$ is greater than 1, the function $y=a^x$ models **exponential growth**. As the input $x$ increases, the output $y$ increases at an ever-accelerating rate. The graph possesses several key, recognisable characteristics:\n\n* It always passes through the point $(0, 1)$, because for any valid base $a$, the zeroth power is defined as $a^0 = 1$.\n* The function's value is always positive, which means the graph lies entirely above the x-axis.\n* The x-axis ($y=0$) serves as a **horizontal asymptote** as $x \\to -\\infty$. The graph approaches the x-axis for large negative values of $x$ but never touches or crosses it.\n* The function is **strictly increasing** for all real values of $x$. For any $x_1 < x_2$, it follows that $a^{x_1} < a^{x_2}$.\n\nFor instance, the graphs of $y=2^x$ and $y=3^x$ both exhibit these features. A larger base results in a steeper graph; for $x>0$, the graph of $y=3^x$ rises more rapidly than $y=2^x$, while for $x<0$, it is closer to the x-axis.\n\n### The Graph of $y = a^x$ for $0 < a < 1$\n\nWhen the base $a$ is between 0 and 1, the function $y=a^x$ models **exponential decay**. As $x$ increases, the value of $y$ decreases, tending towards zero. The principal characteristics of this graph are:\n\n* It also passes through the coordinate point $(0, 1)$, since $a^0 = 1$.\n* The function is always positive, so its graph remains entirely above the x-axis.\n* The x-axis ($y=0$) acts as a **horizontal asymptote**, but this time as $x \\to \\infty$.\n* The function is **strictly decreasing** for all real $x$. For any $x_1 < x_2$, it follows that $a^{x_1} > a^{x_2}$.\n\nIt is useful to recognise that a decay function can be expressed in terms of growth. A base $a$ where $0 < a < 1$ can be written as $1/b$ for some $b>1$. Thus, $y = a^x = (1/b)^x = (b^{-1})^x = b^{-x}$. This shows that the graph of an exponential decay function like $y=(1/2)^x$ is simply a reflection of the growth function $y=2^x$ in the y-axis.", "tips": "When an equation involves terms like $a^{2x}$, $a^x$, and constants, immediately look for a 'disguised' quadratic by making a substitution such as $u = a^x$.", "cautions": "Remember that $a^x$ is always positive for any real $x$ and any positive base $a$. This is a crucial constraint when solving equations, as a substitution like $u = a^x$ implies $u > 0$.", "insights": "The exponential function $y=a^x$ and the logarithmic function $y=\\log_a x$ are inverses of each other. Their graphs are reflections in the line $y=x$, and their properties, such as domain and range, are interchanged.", "examples": [ { "title": "Example: Finding the Maximum Value of an Exponential Function", "question": "Find the maximum value of $4^{\\sin x} - 4 \\times 2^{\\sin x} + \\frac{17}{4}$ for real $x$.", "solution": "### Step-by-Step Solution\n\n**Step 1: Recognise the structure of the expression.**\nThe expression given is $4^{\\sin x} - 4 \\times 2^{\\sin x} + \\frac{17}{4}$. The key insight is to notice that the terms are related by powers of $2$. Specifically, we can write $4^{\\sin x}$ as $(2^2)^{\\sin x} = (2^{\\sin x})^2$. This structure strongly suggests that the expression is a quadratic function in terms of $2^{\\sin x}$.\n\n**Step 2: Make a substitution to form a quadratic.**\nLet's define a new variable, $u = 2^{\\sin x}$. Substituting this into the expression gives us a standard quadratic function of $u$:\n\n$f(u) = u^2 - 4u + \\frac{17}{4}$\n\n**Step 3: Determine the valid range for the new variable $u$.**\nThe variable $x$ can be any real number. The sine function has a range of $-1 \\le \\sin x \\le 1$. Since the exponential function $y=2^t$ is an increasing function, the range of $u = 2^{\\sin x}$ is bounded by $2^{-1}$ and $2^1$. Therefore, the domain for our quadratic in $u$ is $\\frac{1}{2} \\le u \\le 2$.\n\n**Step 4: Find the maximum value of the quadratic on this interval.**\nTo analyse the quadratic $f(u) = u^2 - 4u + \\frac{17}{4}$, we can complete the square. This will reveal the vertex and the shape of the parabola.\n\n$f(u) = (u^2 - 4u + 4) - 4 + \\frac{17}{4}$\n$f(u) = (u - 2)^2 - \\frac{16}{4} + \\frac{17}{4}$\n$f(u) = (u - 2)^2 + \\frac{1}{4}$\n\nThe vertex of this parabola is at $(2, \\frac{1}{4})$. Since the coefficient of the $u^2$ term is positive, this vertex represents the minimum value of the function. The parabola opens upwards.\n\n**Step 5: Analyse the function on the restricted domain.**\nWe are looking for the *maximum* value of $f(u)$ on the interval $[\\frac{1}{2}, 2]$. Since the vertex (the minimum) is at $u=2$, the function must be decreasing across the entire interval from $u=\\frac{1}{2}$ to $u=2$. Consequently, the maximum value must occur at the leftmost point of the interval, which is $u = \\frac{1}{2}$.\n\n**Step 6: Calculate the maximum value.**\nSubstitute $u = \\frac{1}{2}$ into the completed square form of the function:\n\nMaximum value = $(\\frac{1}{2} - 2)^2 + \\frac{1}{4}$\n$= (-\\frac{3}{2})^2 + \\frac{1}{4}$\n$= \\frac{9}{4} + \\frac{1}{4}$\n$= \\frac{10}{4} = \\frac{5}{2}$\n\nThus, the maximum value of the original expression is $\\frac{5}{2}$." } ], "practiceQuestions": [ { "title": "Practice Question 1: Transformations of Logarithmic Graphs", "question": "The graph of $y = \\log_{10}x$ is translated in the positive $y$-direction by $2$ units.\\nThis translation is equivalent to a stretch of factor $k$ parallel to the $x$-axis.\\nWhat is the value of $k$?", "hint": "Translate the equation of the transformed graph algebraically. Use the laws of logarithms to rewrite the vertical shift as a change inside the logarithm, which corresponds to a horizontal stretch.", "solution": "### Algebraic Approach\n\nA translation of the graph of $y = f(x)$ by $2$ units in the positive $y$-direction gives the graph of $y = f(x) + 2$. A stretch of factor $k$ parallel to the $x$-axis gives the graph of $y = f\\left(\\frac{x}{k}\\right)$. For these to be the same, we must have $f(x) + 2 = f\\left(\\frac{x}{k}\\right)$ for all valid $x$.\n\nIn our case, $f(x) = \\log_{10} x$, so we require:\n$\\log_{10} x + 2 = \\log_{10}\\left(\\frac{x}{k}\\right)$\n\nUsing the law of logarithms, $\\log_{10} a - \\log_{10} b = \\log_{10}(a/b)$, the right-hand side is $\\log_{10} x - \\log_{10} k$. So we have:\n$\\log_{10} x + 2 = \\log_{10} x - \\log_{10} k$\n\nSubtracting $\\log_{10} x$ from both sides gives:\n$2 = -\\log_{10} k$\n$\\log_{10} k = -2$\n\nBy the definition of logarithms, this means $k = 10^{-2} = \\frac{1}{100} = 0.01$.\n\n### Single Point Approach (x-intercept)\n\nAn alternative method is to track the x-intercept of the graph through the transformations.\n1. **Original graph**: The graph of $y=\\log_{10}x$ has its x-intercept where $y=0$, which is at $x=1$ (the point is $(1,0)$).\n2. **Translated graph**: The new graph is $y' = \\log_{10}x + 2$. Its x-intercept is where $y'=0$. So, $\\log_{10}x + 2 = 0 \\implies \\log_{10}x = -2 \\implies x = 10^{-2} = 0.01$.\n3. **Stretched graph**: The stretched graph is $y'' = \\log_{10}(\\frac{x}{k})$. Its x-intercept is where $y''=0$. So, $\\log_{10}(\\frac{x}{k}) = 0 \\implies \\frac{x}{k} = 1 \\implies x=k$.\n4. **Equivalence**: Since the translated graph and the stretched graph are equivalent, their x-intercepts must be identical. Therefore, $k = 0.01$. The correct answer is A." }, { "title": "Practice Question 2: Identifying Complex Exponential Graphs", "question": "P, Q, R and S show the graphs of\\\\n$y = (\\\\cos x)^{\\\\cos x}$, $y = (\\\\sin x)^{\\\\sin x}$, $y = (\\\\cos x)^{\\\\sin x}$ and $y = (\\\\sin x)^{\\\\cos x}$\\\\nfor $0 < x < \\\\frac{\\\\pi}{2}$ in some order.\\\\n![](https://cdn.ether.sh/images/98c234/f9b5c928297beb3e0f491fe6fcd340fdda28fbce80c1f3fee6c63879d1de2a40.jpg)\\\\nWhich row in the following table correctly identifies the graphs?\n![](https://cdn.ether.sh/images/TMUA/1753206697933_4yma2nsa74k.png)", "hint": "Evaluate each function at key points in the domain $0 < x < \\frac{\\pi}{2}$, such as the limits as $x \\to 0^+$ and $x \\to \\frac{\\pi}{2}^-$, to match them to the graphs.", "solution": "Let's analyse the behaviour of each function at the boundaries of the domain $0 < x < \\frac{\\pi}{2}$. We will consider the limits as $x$ approaches $0$ from the right ($x \\to 0^+$) and as $x$ approaches $\\frac{\\pi}{2}$ from the left ($x \\to \\frac{\\pi}{2}^-$).\n\n### Behaviour as $x \\to 0^+$\n* $y_1 = (\\cos x)^{\\cos x}$: As $x \\to 0^+$, $\\cos x \\to 1$. So, $y_1 \\to 1^1 = 1$.\n* $y_2 = (\\sin x)^{\\sin x}$: As $x \\to 0^+$, $\\sin x \\to 0$. This is a limit of the form $u^u$ as $u \\to 0^+$, which is a standard result evaluating to $1$. So, $y_2 \\to 1$.\n* $y_3 = (\\cos x)^{\\sin x}$: As $x \\to 0^+$, $\\cos x \\to 1$ and $\\sin x \\to 0$. So, $y_3 \\to 1^0 = 1$.\n* $y_4 = (\\sin x)^{\\cos x}$: As $x \\to 0^+$, $\\sin x \\to 0$ and $\\cos x \\to 1$. So, $y_4 \\to 0^1 = 0$.\n\nFrom the graphs, only graph Q approaches $0$ as $x \\to 0^+$. Therefore, **Graph Q is $y = (\\sin x)^{\\cos x}$**.\n\n### Behaviour as $x \\to \\frac{\\pi}{2}^-$\n* $y_1 = (\\cos x)^{\\cos x}$: As $x \\to \\frac{\\pi}{2}^-$, $\\cos x \\to 0$. Limit is of the form $u^u$ as $u \\to 0^+$, so $y_1 \\to 1$.\n* $y_2 = (\\sin x)^{\\sin x}$: As $x \\to \\frac{\\pi}{2}^-$, $\\sin x \\to 1$. So, $y_2 \\to 1^1 = 1$.\n* $y_3 = (\\cos x)^{\\sin x}$: As $x \\to \\frac{\\pi}{2}^-$, $\\cos x \\to 0$ and $\\sin x \\to 1$. So, $y_3 \\to 0^1 = 0$.\n* $y_4 = (\\sin x)^{\\cos x}$: As $x \\to \\frac{\\pi}{2}^-$, $\\sin x \\to 1$ and $\\cos x \\to 0$. So, $y_4 \\to 1^0 = 1$.\n\nFrom the graphs, only graph P approaches 0 as $x$ approaches $\\frac{\\pi}{2}$. Therefore, **Graph P is $y = (\\cos x)^{\\sin x}$**.\n\n### Distinguishing R and S\nWe are left with $y = (\\cos x)^{\\cos x}$ and $y = (\\sin x)^{\\sin x}$ for graphs R and S. Both of these functions are of the form $f(u) = u^u$, where $u$ is either $\\cos x$ or $\\sin x$. The function $f(u) = u^u$ has a minimum value at $u = \\frac{1}{e} \\approx 0.37$.\n\nFor $y = (\\sin x)^{\\sin x}$, the minimum will occur when $\\sin x = \\frac{1}{e}$. Since $\\sin x$ is an increasing function on $(0, \\frac{\\pi}{2})$, this happens at a relatively small value of $x$.\n\nFor $y = (\\cos x)^{\\cos x}$, the minimum will occur when $\\cos x = \\frac{1}{e}$. Since $\\cos x$ is a decreasing function on $(0, \\frac{\\pi}{2})$, this happens at a value of $x$ closer to $\\frac{\\pi}{2}$.\n\nComparing the graphs, Graph S has its minimum at a smaller value of $x$ than Graph R. Therefore, **S is $y = (\\sin x)^{\\sin x}$** and **R is $y = (\\cos x)^{\\cos x}$**.\n\nMatching these gives the row:\n$y = (\\cos x)^{\\cos x}$: R\n$y = (\\sin x)^{\\sin x}$: S\n$y = (\\cos x)^{\\sin x}$: P\n$y = (\\sin x)^{\\cos x}$: Q\nThis corresponds to option E." }, { "title": "Practice Question 3: Exponential Inequalities", "question": "Three real numbers $x, y$ and $z$ satisfy $x > y > z > 1$.\nWhich one of the following statements must be true?", "hint": "For each inequality, try to find a counterexample using values that satisfy $x > y > z > 1$. If you cannot find a counterexample, try to prove why the statement must be true by manipulating the inequality using index laws.", "solution": "The question asks which statement **must** be true given $x > y > z > 1$. This suggests we can test each option, trying to find a counterexample for those that are not always true.\n\n**A: $\\\\frac{2^{z+1}}{2^x} > \\\\frac{2^x + 2^z}{2^y}$**\nThis simplifies to $2^{z-x+1} > 2^{x-y} + 2^{z-y}$. Since $x>z$, $z-x$ is negative. If $x$ is much larger than $z$, $z-x+1$ will be a large negative number, making the left side very small. Since $x>y$, $x-y$ is positive, making $2^{x-y}$ potentially large. Let $x=10, y=5, z=2$. LHS is $2^{2-10+1} = 2^{-7} = 1/128$. RHS is $2^{10-5} + 2^{2-5} = 2^5 + 2^{-3} = 32 + 1/8$. Clearly $1/128 > 32.125$ is false. So A is not always true.\n\n**B: $2 > \\\\frac{3^x + 3^z}{3^y}$**\nThis simplifies to $2 > 3^{x-y} + 3^{z-y}$. Since $x>y$, the term $3^{x-y}$ can be made arbitrarily large by choosing $x$ much larger than $y$. For example, let $x=4, y=2, z=1.5$. RHS is $3^{4-2} + 3^{1.5-2} = 3^2 + 3^{-0.5} = 9 + 1/\\sqrt{3}$. This is much greater than 2. So B is not always true.\n\n**C: $\\\\frac{2 \\\\times 5^x}{5^z} > \\\\frac{5^x + 5^z}{5^y}$**\nLet's clear the denominators by multiplying by $5^y \\times 5^z$, which is positive. The inequality becomes $2 \\times 5^x \\times 5^y > (5^x + 5^z) \\times 5^z$, which simplifies to $2 \\times 5^{x+y} > 5^{x+z} + 5^{2z}$.\nLet's prove this must be true. We can write the LHS as $5^{x+y} + 5^{x+y}$. We need to show that this is greater than $5^{x+z} + 5^{2z}$. We can do this by comparing terms.\n1. Compare $5^{x+y}$ with $5^{x+z}$. Since $x>y>z>1$, we have $y > z$, so $x+y > x+z$. As the base is $5>1$, this implies $5^{x+y} > 5^{x+z}$.\n2. Compare $5^{x+y}$ with $5^{2z}$. We have $x > z$ and $y > z$. Adding these inequalities gives $x+y > 2z$. Since the base is $5>1$, this implies $5^{x+y} > 5^{2z}$.\nSince $5^{x+y}$ is strictly greater than both terms on the RHS, their sum must be less than $5^{x+y} + 5^{x+y} = 2 \\times 5^{x+y}$. So the inequality must be true.\n\n**D: $2 < \\\\frac{7^x + 7^z}{7^y}$**\nThis simplifies to $2 < 7^{x-y} + 7^{z-y}$. Since $x>y$, $x-y>0$, so $7^{x-y} > 1$. Since $y>z$, $z-y<0$, so $0 < 7^{z-y} < 1$. It is possible for the sum to be less than 2. Let $y$ be very close to $x$, and $z$ be much smaller than $y$. For example, let $x=10, y=9.99, z=2$. Then $x-y = 0.01$ and $z-y = -7.99$. The RHS is $7^{0.01} + 7^{-7.99}$. $7^{0.01}$ is only slightly greater than 1 (as any positive number to the power of 0 is 1), and $7^{-7.99}$ is a very small positive number. Their sum will be slightly greater than 1, and certainly can be less than 2. So D is not always true.\n\nTherefore, only statement C must be true." } ] }