Given the graph of y = f'(x), determine: (a) the gradient of the tangent to f(x) at x=0; (b) the value of x where f(x) has a tangent parallel to the tangent at x=0; (c) the intervals where f(x) is dec

The gradient of the tangent to the graph of $f(x)$ is given by its derivative, $f'(x)$. We need to find the value of $f'(x)$ at $x=0$ from the given graph of $y=f'(x)$.
From the graph, when $x=0$, the value of $y=f'(x)$ is 2.

The gradient of the tangent to the graph of $f(x)$ at $x=0$ is $\boxed{2}$.

A tangent to the curve $f(x)$ will be parallel to the tangent in (a) if it has the same gradient, which is 2. We need to find another value of $x$ where $f'(x)=2$.
Looking at the graph, the parabola $y=f'(x)$ is symmetric. The $x$-intercepts are at $x=-1$ and $x=5$, so the axis of symmetry is at $x = \frac{-1+5}{2} = 2$.
Since $f'(0)=2$, by symmetry, $f'(x)$ will also be 2 at $x = 2 + (2-0) = 4$.
So, $f'(4)=2$.
The value of $x$ for which there will be a tangent to the curve $f(x)$ parallel to the tangent in (a) is $\boxed{4}$.

The function $f(x)$ is decreasing when its derivative $f'(x)$ is negative ($f'(x) < 0$).
From the graph, $f'(x)$ is below the x-axis for $x < -1$ and for $x > 5$.
So, $f(x)$ is decreasing for $\boxed{x < -1 \text{ or } x > 5}$.

Turning points of $f(x)$ occur where $f'(x) = 0$.
From the graph, $f'(x)=0$ at $x=-1$ and $x=5$.

• At $x=-1$: $f'(x)$ changes from negative to positive. This indicates a local minimum.
• At $x=5$: $f'(x)$ changes from positive to negative. This indicates a local maximum. The $x$-coordinates of the turning points of $f(x)$ are $\boxed{x=-1 \text{ (local minimum)} \text{ and } x=5 \text{ (local maximum)}}$.

To draw the graph of $f(x)$, we use the following information:

• $x$-intercepts of $f(x)$: $-2, 2, 7$.
• Local minimum at $x=-1$.
• Local maximum at $x=5$.
• $f(x)$ is decreasing for $x < -1$.
• $f(x)$ is increasing for $-1 < x < 5$.
• $f(x)$ is decreasing for $x > 5$.

Based on this, the graph of $f(x)$ would be a cubic function with the following characteristics:

1. It starts from positive infinity, decreasing until it crosses the x-axis at $x=-2$.
2. It continues decreasing to reach a local minimum at $x=-1$.
3. It then increases, crossing the x-axis at $x=2$.
4. It continues increasing to reach a local maximum at $x=5$.
5. Finally, it decreases, crossing the x-axis at $x=7$ and continuing towards negative infinity.

The sketch would show a curve that starts high, dips to a minimum, rises to a maximum, and then falls. The $x$-intercepts and turning points are consistent with this shape.