Question - Finding the Derivative of a Function Expression
Solution:
The image shows two graphs, y = f(x) and y = g(x), intersecting on a Cartesian coordinate plane, and there are lines $$ d_1 $$ and $$ d_2 $$ which appear to be parallel to the x-axis at different y-values. The function $$ h(x) $$ is defined such that $$ h(x + 1) = f(x) - g(x + 2) $$. We are tasked with finding the value of $$ h'(2) $$.The first step to find $$ h'(2) $$ is to differentiate the equation for $$ h(x + 1) $$ with respect to $$ x $$. So, let's differentiate $$ h(x + 1) $$ using the chain rule:If $$ u = x + 1 $$, then $$ du/dx = 1 $$ and so,\[ h'(u) = \frac{dh}{du} \cdot \frac{du}{dx} = \frac{dh}{du} \]Now differentiate $$ f(x) $$ and $$ g(x + 2) $$ separately with respect to $$ x $$. Using the property $$ \frac{d}{dx}f(x) = f'(x) $$ and $$ \frac{d}{dx}g(x+2) $$ with the chain rule ($$ v = x + 2 $$, $$ dv/dx = 1 $$), we get:\[ h'(u) = f'(x) - g'(v) \]Knowing that $$ h'(u) = h'(x + 1) $$, we can convert the differentiation point from $$ x $$ to $$ u $$ to find $$ h'(2) $$:\[ h'(2) = f'(1) - g'(4) \]To get the values of $$ f'(1) $$ and $$ g'(4) $$, we need to find the slopes of $$ f(x) $$ and $$ g(x) $$ at $$ x = 1 $$ and $$ x = 4 $$, respectively.Looking at the graph, $$ f(x) $$ passes through point A(1,0), and since $$ d_1 $$ and $$ d_2 $$ are parallel to the x-axis, it means the slope is 0 at $$ x = 1 $$ for $$ f(x) $$. Therefore, the slope $$ f'(1) = 0 $$.The graph doesn't give us $$ g(4) $$ directly, but we can use the fact that the line through the point B(-1, 2) is at a 45-degree angle to infer that this line, representing $$ g(x) $$, has a slope of 1, as a 45-degree line in the first quadrant will have a slope of 1. Hence, $$ g'(x) = 1 $$ for all $$ x $$ in the domain of $$ g $$, and specifically $$ g'(4) = 1 $$.Now, substitute the slope values into the equation for $$ h'(2) $$:\[ h'(2) = 0 - 1 = -1 \]Thus, the value of $$ h'(2) $$ is $$-1$$.
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