{"id":2515,"date":"2013-05-01T15:34:27","date_gmt":"2013-05-01T15:34:27","guid":{"rendered":"http:\/\/testpreparations.com\/help\/?p=2515"},"modified":"2014-12-02T08:27:02","modified_gmt":"2014-12-02T08:27:02","slug":"math-review-composition-of-functions","status":"publish","type":"post","link":"https:\/\/schooltutoring.com\/scholarship\/2013\/05\/01\/math-review-composition-of-functions\/","title":{"rendered":"Math Review: Composition of Functions"},"content":{"rendered":"

By now, you have probably dealt with a cornucopia of functions with two variables; most commonly x<\/em> and y<\/em>. To find points that exist within the function you would substitute values into one variable in order to find the other, for example subbing an x<\/em> value into a linear equation to find it’s corresponding y<\/em> component. Not only can you sub numeric values into equations, but entire equations as well. The composition of two functions, say f<\/em> and g<\/em>, creates a new functions. This function is solved by performing f<\/em> and then performing g<\/em>.<\/p>\n

For example, consider two functions: g(x) = x^3<\/em> and f(x) = x – 4.<\/em> The composition of f with g is called f\u25cbg<\/em> and is worked out as<\/p>\n

f\u25cbg = f(g(x))<\/em><\/p>\n

First we write down what g(x)<\/em> is followed by applying f(x)<\/em> to the whole of g(x)<\/em>. In this case, applying f(x)<\/em> means subtracting four.<\/p>\n

f(g(x)) = (x^3) \u2013 4 = x^3 \u2013 4<\/em><\/p>\n

Another example would be if we considered the composition of functions h(x) = x^2 + x +2<\/em> with j(x) = e^x<\/em>. Just as we did previously, we will write out j<\/em> and then apply h<\/em> to the whole of j<\/em>.<\/p>\n

h\u25cbj = h(j(x)) = (e^x)^2 + e^x + 2 = e^2x + e^x + 2<\/em><\/p>\n

The order in which we compose functions makes a difference in what our composite function will be. Using f(x)<\/em> and g(x)<\/em> from the first example, we can see that if we reverse the order in which we compose them we get drastically different functions.<\/p>\n

g\u25cbf = g(f(x)) = (x-4)^3 = x^3 \u2013 12x^2 + 48x -64<\/em><\/p>\n

As you can see, this function is very different from f(g(x))<\/em>. In general, f(g(x))<\/em> is not equal to g(f(x))<\/em>.<\/p>\n

Occasionally, you may be presented with a function that is already a composition of two functions and asked to find these two functions. This process is known as decomposition. Let’s look at the following function and consider it: h(x) = sin(4x)<\/em>. Based on our knowledge of composite functions, we can see that h(x)<\/em> can be written as f(g(x))<\/em>. We know we start by writing what g(x)<\/em> is first, followed by applying f<\/em> to the whole of g(x)<\/em>.<\/p>\n

h(x) = sin(4x) = f(4x) = f(g(x))<\/em><\/p>\n

The above is the reverse order of composing functions and shows that g(x) = 4x<\/em> and f(x) = sin(x).<\/em><\/p>\n

One thing to remember when solving problems with composite functions is that not all functions can be composed together where as some can only be composed for a certain set of x values. This condition is determined by the domain of both functions. The domain of a composed function is either the domain of the first function, or it lies inside the domain of the first function. Similarly, the range of a composed function is either the range of the second function, or else is inside it.<\/p>\n

Looking to do the PSAT? We can help with PSAT Prep<\/a><\/p>\n

This article was written for you by Troy<\/strong>, one of the tutors with Test Prep Academy<\/span>.<\/p>\n","protected":false},"excerpt":{"rendered":"

By now, you have probably dealt with a cornucopia of functions with two variables; most commonly x and y. To find points that exist within the function you would substitute values into one variable in order to find the other, for example subbing an x value into a linear equation to find it’s corresponding y […]<\/p>\n","protected":false},"author":6,"featured_media":7811,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"inline_featured_image":false,"footnotes":""},"categories":[2841,3015,3021,2851],"tags":[3139,3140,3159,3349],"class_list":["post-2515","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-act","category-math-act","category-mathematics-sat","category-sat","tag-how-to-compose-functions","tag-how-to-decompose-functions","tag-how-to-solve-composite-functions","tag-what-are-composite-functions"],"acf":[],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/posts\/2515","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/users\/6"}],"replies":[{"embeddable":true,"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/comments?post=2515"}],"version-history":[{"count":0,"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/posts\/2515\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/media\/7811"}],"wp:attachment":[{"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/media?parent=2515"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/categories?post=2515"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/tags?post=2515"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}