Example Question - function composition
Here are examples of questions we've helped users solve.
Analyzing the Composition of a Given Function Graph
Given the function \( f(x) = 2x^3 - 9x^2 + 12x - 3 \), we are asked to determine its concavity, maximum point, and inflection point. To find the concavity of the function, we need to compute the second derivative and analyze its sign: \( f''(x) = \frac{d^2}{dx^2}(2x^3 - 9x^2 + 12x - 3) \) \( f''(x) = \frac{d}{dx}(6x^2 - 18x + 12) \) \( f''(x) = 12x - 18 \) To determine the concavity, we look at the sign of \( f''(x) \). If \( f''(x) > 0 \), the function is concave up. If \( f''(x) < 0 \), the function is concave down. So, the function is concave up when \( 12x - 18 > 0 \) i.e., \( x > \frac{3}{2} \), and concave down when \( x < \frac{3}{2} \). To find the maximum point, we need to set the first derivative equal to zero and solve for x: \( f'(x) = \frac{d}{dx}(2x^3 - 9x^2 + 12x - 3) \) \( f'(x) = 6x^2 - 18x + 12 \) Setting \( f'(x) = 0 \), we get: \( 6x^2 - 18x + 12 = 0 \) Dividing by 6: \( x^2 - 3x + 2 = 0 \) Factoring: \( (x - 2)(x - 1) = 0 \) Thus, \( x = 1 \) and \( x = 2 \) are critical points. We must check these points in the second derivative to see if they correspond to maximum points: \( f''(1) = 12(1) - 18 = -6 < 0 \), so \( x = 1 \) is a local maximum point. \( f''(2) = 12(2) - 18 = 6 > 0 \), so \( x = 2 \) is not a maximum point, it is a point of inflection since the concavity changes from down to up.
Finding the Composition Function
The question asks you to find a function \( g(x) \) such that the composition of functions \( h(g(x)) \) is equal to the function \( f(x) \) provided. Here are the given functions: \( h(a) = \frac{1}{\sqrt{2a-1}} \) \( f(x) = x + 2 \) We want to find \( g(x) \) such that when we plug \( g(x) \) into \( h(a) \), we get \( f(x) \). So, we are solving for \( g(x) \) in the equation: \( h(g(x)) = f(x) \) \( \frac{1}{\sqrt{2g(x)-1}} = x + 2 \) Let's solve for \( g(x) \) step by step: 1. Get rid of the fraction by multiplying both sides by \( \sqrt{2g(x)-1} \): \( \sqrt{2g(x)-1} = \frac{1}{x + 2} \) 2. Square both sides to get rid of the square root: \( 2g(x)-1 = \frac{1}{(x + 2)^2} \) 3. Isolate \( g(x) \) by adding 1 to both sides: \( 2g(x) = \frac{1}{(x + 2)^2} + 1 \) 4. Divide both sides by 2 to solve for \( g(x) \): \( g(x) = \frac{1}{2(x + 2)^2} + \frac{1}{2} \) Let's simplify the equation for \( g(x) \): First, find a common denominator for the addition: \( g(x) = \frac{1 + (x + 2)^2}{2(x + 2)^2} \) Now let's expand \( (x + 2)^2 \): \( (x + 2)^2 = x^2 + 4x + 4 \) Substitute this back into \( g(x) \): \( g(x) = \frac{1 + x^2 + 4x + 4}{2(x^2 + 4x + 4)} \) This further simplifies to: \( g(x) = \frac{x^2 + 4x + 5}{2(x^2 + 4x + 4)} \) So the function \( g(x) \) that meets the requirement \( h(g(x)) = f(x) \) is: \( g(x) = \frac{x^2 + 4x + 5}{2(x^2 + 4x + 4)} \)
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