1. The problem states that we have a function $y = f(x)$ and another function defined as $g(x) = x + f(x)$. We are asked to find the graph of the derivative $y = g'(x)$. 2. Recall the rule for derivatives: the derivative of a sum is the sum of the derivatives. So, $$g'(x) = \frac{d}{dx}[x + f(x)] = \frac{d}{dx}[x] + \frac{d}{dx}[f(x)] = 1 + f'(x).$$ 3. This means the graph of $g'(x)$ is the graph of $f'(x)$ shifted upward by 1 unit. 4. To understand $g'(x)$, we need to analyze $f'(x)$, the derivative of $f(x)$. The derivative $f'(x)$ represents the slope of the tangent line to the curve $y=f(x)$ at each point. 5. From the graph of $f(x)$: - At $x=-2$, the curve is increasing (slope positive). - At $x=-1$, the curve reaches a peak (slope $f'(-1)=0$). - Between $-1$ and $0$, the curve is decreasing (slope negative). - At $x=0$, the curve crosses the origin going downward (slope negative). - At $x=1$, the curve reaches a valley (slope $f'(1)=0$). - After $x=1$, the curve is increasing sharply (slope positive). 6. Therefore, $f'(x)$ is positive before $x=-1$, zero at $x=-1$, negative between $-1$ and $1$, zero at $x=1$, and positive after $1$. 7. Since $g'(x) = 1 + f'(x)$, the graph of $g'(x)$ is the graph of $f'(x)$ shifted up by 1. This means: - Where $f'(x)$ was zero, $g'(x)$ will be 1. - Where $f'(x)$ was positive, $g'(x)$ will be greater than 1. - Where $f'(x)$ was negative, $g'(x)$ will be less than 1. 8. In summary, the graph of $y = g'(x)$ looks like the slope graph of $f(x)$ shifted up by 1 unit. Final answer: $$g'(x) = 1 + f'(x).$$