Posts tagged with: functions

I’m having a lot of trouble with this problem, could you please break it down for me? It is number 2 on the functions practice questions. “If f(x-1)=x+1 for all values of x, which of the following is equal to f(x+1)?”
A. x+3
B. x+2
C. x-1
D. x-3

In the solutions in the back of the book, I have two approaches. I’m assuming you’ve already looked at those and neither went all the way for you, so here’s another look.

We know that function notation means, generally, that the bit in parentheses goes through some series of mathematical operations, and a result is spit out. So, for example, stuff can be added or subtracted, or the argument could be multiplied, divided, raised to a power, etc.

In this particular problem, we know that f(x-1) goes through some process that spits out an end result of x+1. Here’s the crux: what could happen to (x-1) to make it equal x+1? No multiplying or raising to powers necessary! The only thing you need to do to turn (x-1) into x+1 is add 2!

That’s how we know that the function f just adds 2.



Therefore, we can figure out what f(x+1) must be:


Does that help?

The way to think about this (for me, anyway) begins with understanding that 2[something] + 3 = 8x – 1, and our job is to figure out what that something is.

Since this question gives us answer choices, all we really need to do is try each one as the something to see what works. Since they’re not in numerical order, start with A.

2(4x – 2) + 3
8x – 4 + 3
8x – 1

Oooh, look at that! we’re already done! 🙂

(Also, for those worrying along at home, note that 5 answer choices means this isn’t an SAT question. Could be SAT Subject Test, though.)

If you didn’t have answer choices, you’d still start in the same place, but then think through logically. Thought process:

2[something] + 3 = 8x – 1

Hmm..there’s gotta be a 4x in the something otherwise i’ll never get 8x. But if it’s just 4x in there then I’ll end up with 8x + 3, not 8x – 1. How do i take 4 away? By adding a –2 inside the something!

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Think of it this way: the g function is doing SOME AS-YET-UNKNOWN THINGS to (–x + 7) to turn it into (2x + 1). Of the simple mathematical operations probably at play here (addition, subtraction, multiplication, division) what could be going on?

First, the only way you go from –x to 2x is you multiply by –2. So let’s see what happens if we just multiply f(x) by –2.

–2(–x + 7) = 2x – 14

OK, so the first part’s good now, but how can we turn –14 into +1? Well, we don’t want to multiply or divide again because that would screw up the 2x we just nailed down, so why don’t we try adding 15?

2x – 14 + 15 = 2x + 1

Combine the two operations we just did (multiply by –2, add 15) and you have the g function. The function g will multiply its argument by –2, then add 15. Mathematically, we can write that like this:

g(x) = –2x + 15

Now, start from the top and make sure we’re right.

= g(–x + 7)             <– substitute (–x + 7) for f(x)
= –2(–x + 7) + 15   <– apply the g function to (–x + 7)
= 2x – 14 + 15
= 2x + 1

It works! Now all we need to do is calculate g(2).

g(2) = –2(2) + 15
g(2) = 11

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The thing to remember about functions is that they do the same thing to whatever is inside the parentheses. So don’t worry about the r vs. the h. They could use x, or a little star symbol, or whatever else they want. What matters is that the function f, as defined here, will equal zero when r = 4, or when r = –1.

When we’re told that f(h – 3) = 0, we can conclude that h – 3 must equal either 4 or –1. Therefore, h must equal either 7 or 2.

The graph below might help show what’s going on visually. When the thing in the parentheses next to the f (also known as the argument of the function) equals –1 or 4, the function equals zero. Therefore h – 3 must equal –1 or 4.

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A function will only have that property if it’s a line that passes through the origin. For example, f(x) = 5x has that property:

You can try the same with other linear functions to see why they won’t work. For example, if f(x) = 5x + 2:

Nonlinear functions also won’t work. For example, if f(x) = x^2:

Anyway, now that we know we’re dealing with a linear function through the origin, we can figure out that if f(6) = 12, then the function we’re dealing with must be f(x) = 2x. Therefore, f(2) = 4.

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Thank you, Mike, for your ever-awesome explanations! Here’s a question from April 2017 SAT, Section 4.22:

The graphs in the xy-plane of the following quadratic equations each have x-intercepts of -2 and 4. The graph of which equation has its vertex farthest from the x-axis?

A) y = -7 (x + 2)(x – 4)

B) y = 1/10 (x + 2)(x – 4)

C) y = -1/2 (x + 2)(x – 4)

D) y = 5 (x + 2)(x – 4)

It’s useful to note that all these equations are the same after the leading coefficient. So, for example, we could just say f(x)=(x+2)(x-4) and then the choices would just be:

A) y=-7f(x)

B) y=\frac{1}{10}f(x)

C) y=-\frac{1}{2}f(x)

D) y=5f(x)

The reason I point that out is that the effect of multiplying a function by a constant, in general, is that the larger the absolute value of the constant is, the more you’ll be stretching that function away from the x-axis. If you know that little fact, then you can immediately pick the answer choice with the coefficient with the largest absolute value, which is choice A.

If you don’t know that little fact, but you have a graphing calculator, you’re still in good shape. Just graph each one (on the same screen, if possible) to see which one has its vertex farthest from the x-axis.

Note that f(x) is a line in slope-intercept form, where a is the slope and b is the intercept.

Once you recognize that, you just need to know that, notationally, the –2 in front of f(x) means you multiply the whole thing by –2. So:


That’s still a line, but now the slope is –2a.

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Note that f(x) is a line in slope-intercept form, where a is the slope and b is the intercept.

Once you recognize that, you just need to know that, notationally, the –2 in front of f(x) means you multiply the whole thing by –2. So:


That’s still a line, but now the slope is –2a.

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A question like this on the SAT will always have answer choices, and in this case would be easy to backsolve. Don’t rob yourself of easy points—remember that answer choices are often a helpful tool! Use them to your advantage. In this case, you’d simply need to plug (2x – 1) in for x in each answer choice until one gave you cx.

Since you didn’t provide answer choices, though, we’ll have to do this the hard way.

The way I like to think about questions like this is pretty procedural. You’ve got some f(x) expression. You swap out the x for (2x – 1) and what happens? First, the 2x means there’s a doubling. We know we want to land on cx, so f(x) must have a c/2 in the x term that gets doubled. Make sense?


But what about those question marks? Well, we know that the end result we want is that f(2x – 1) = cx, so that question mark must cancel out the –c/2 we get from multiplying c/2 by (2x – 1). What cancels out –c/2? Easy: +c/2! So let’s see if making f(x) = (c/2)x + c/2 works:


Yep, that works. Again, it probably would have been easier to just start with the answer choices and see which one landed you on cx. Your work would end up looking just like that second bit of math above, but without so much abstract thinking required.

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If f and g are functions, where f(x) = x^3 -10x^2 +27x – 18 and g(x) = x^3 – x^2 – 6x, which of the following gives a relationship between f and g?

A) g(x) = 3 f(x)
B) g(x) = f(x) – 3
C) g(x) = f(x) + 3
D) g(x) = f(x – 3)
E) g(x) = f(x + 3)

Can you solve w/o graphing?


Note that the first term in both functions is x^3, so the relationship isn’t multiplication. Eliminate choice A.

Note also that it’s obvious that g(x) isn’t just 3 bigger or 3 smaller than f(x), so you can also eliminate choices B and C without any real work.

From there, it gets a little tricky. Focus on the constant terms, though. Note that f(x) has a –18 in it that goes away in g(x). Note also that when we plug x-3 or x+3 in for x, we’ll have a bunch of binomials to expand, but the only way that –18 goes away is if those binomial expansions spit out a positive 18. Without doing all the math, you should be confident that choice E is the one that will do that. To test it quickly, though, set x=0 (since this does need to be true when x=0).

    \begin{align*}g(0)&=f(0+3)\\g(0)&=f(3)\\0^3 - 0^2 - 6(0)&=3^3 -10(3)^2 +27(3) - 18\\0&=27-90+81-18\\0&=0\end{align*}

Yep! That worked nicely.

Could you please explain SAT Past paper 3, section 4, question 12? I don’t really understand why the double root is considered a ‘distinct’ zero.

Thank you for your time!

Sure. Would it make you feel better if “distinct” were replaced with “unique”? Those terms are interchangeable here.

When the question says the function has “five distinct zeros,” that really just means there are five values that 1) don’t equal each other, and 2) make the function equal zero.

In other words, f(a) = 0, f(b) = 0, f(c) = 0, f(d) = 0, and f(e) = 0 for unique values abcd, and e.

In OTHER other words, the function f touches the x-axis five times. Sidestepping all of the above, all you really need to do is count the number of times each graph touches the x-axis.

Choices A and B touch the x-axis four times, and choice C touches the x-axis six times. Only choice D touches it five times.

Hi Mike, Can you please explain Question 11, Test 6, section 3 ? I know the parabola opens downward, but I’m confused after that. Thanks.

You’ve already got the first part—when you add a negative to the front of a parabola you’ll flip it vertically. Since this parabola begins facing up, the negative flips it so it faces down.

From there, I think it’s helpful to think about this in terms of how functions shift. For some function f(x) that’s graphed on the xy-plane:

  • f(x)+1 ⇒ (graph moves UP one)
  • f(x)–1 ⇒ (graph moves DOWN one)
  • f(x+1) ⇒ (graph moves LEFT one)
  • f(x–1) ⇒ (graph moves RIGHT one)

That applies here. You’re basically taking the ax^2 function and applying a few shifts (and a flip). Check it out–if you start by assuming a function p(x)=ax^2, you can build the function in this question by shifting that original function:


-p(x-b)+c flips the function p(x) upside-down, then shifts it b units to the right and c units up.

If you hate everything I’ve just said, I have good news! You can also just memorize the vertex form of a parabola (which we basically just derived).

If you have a parabola in the form f(x)=a(x-h)^2+k, then you know it has its vertex at (h, k) and that the sign of a tells you whether the parabola opens up or down. This question basically gives you the vertex form, only it uses b and c instead of h and k. Recognize that and you know right away that the parabola’s vertex is at (b, c).

I don’t understand a question from your book. Question #2 on functions. Could you explain it in more detail?

The question:

I like to approach this by thinking about functions abstractly. A function is a process by which the thing in the parenthesis (AKA the argument) is transformed into something else using only mathematical operations. In this case, for all values of x, some mathematical operation or operations are done to x-1 to transform it into x+1.

What could those operations be? You can rule out multiplication or division—the leading coefficient of both x-1 and x+1 is 1. Likewise, you can easily rule out exponents or roots. But what about addition/subtraction? Could something be added to or subtracted from x-1 to transform it into x+1? Sure! If you add 2 to x-1, you get x+1.

So let’s think about how that works in function notation. If we want to say that the function adds 2 to its input, we can write:


That tells us what f(x) would be:


And of course from there, we can see what f(x+1) would be:


Does that help?

Test 8 Section 3 #15

You’re given:


Then you’re asked for the value of h(0). In function-in-function questions like this, the thing to remember is that you plug in the argument for every appearance of x in a function, including other functions in that function. In other words, because we know that h(x)=1-g(x), we know that h(0)=1-g(0).

And of course, from the first function we’re given, we can calculate the value of g(0):


Now plug that back into the equation for h(0):



Functions f and g intersect at the points (–3,5), (2,8) and (11,–17). Which of the following could define the function h(x) = f(x) – g(x)?

A) h(x) = (x + 3)(x – 2)(x –11)
B) h(x) = (x – 5)(x – 8)(x + 17)
C) h(x) = (x + 5)(x – 6)(x + 6)
D) h(x) = (x – 3)(x + 2)(x + 11)

The idea here—and it’s a tricky one—is that when you add or subtract functions, you’re doing that to the y-values for each x-value. For example, let’s say we have functions a and b, and that function a contains points (1, 3) and (2, –1) while function b contains points (1, 1) and (2, 6). If we added a(x) and b(x), the resulting function would contain points (1, 4) and (2, 5)—we sum the y-values at each x-value. Hopefully the illustration using linear functions below isn’t too noisy to look at.

In the above, a(x)=-4x+7 and b(x)=5x-4 contain the points named above for functions a and b. The sum of those is a(x)+b(x)=-4x+7+5x-4=x+3, and it contains (1, 4) and (2, 5).

With that understanding we can solve this problem. When we know that two functions intersect at a point, then we know their y-values are equal at that point. Therefore, we know that if we subtract one intersecting function from the other, the resulting function should have a y-value of zero at that x-value. In other words, if functions f and g intersect at (–3, 5), (2, 8) and (11, –17), we should expect h(x)=f(x)-g(x) to contain the points (–3, 0), (2, 0), and (11, 0).

Choice A does that for us. It will have x-intercepts at –3, 2, and 11.

One final note about the answer choices, which of course are always a potential aid in a multiple choice test. Looking at the choices here, all are in factored form to highlight their x-intercepts, and 3 of them are built up from numbers in the intersection points we’re given. That alone won’t get you all the way there, but might get you thinking along the right lines. When you’re stumped, look to the answer choices for a little inspiration.