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Draw sketches of possible graphs for which the following data hold. Consider only the domain 0x50 \leqslant x \leqslant 5, and assume that the graph of y=f(x)y=f^{\prime \prime}(x) is smooth. (a) f(0) = 0, f(2) = 5, f'(2) = 0, f"(2) < 0, f(4) = 3, f'(4) = 0, f"(4) > 0 (b) f(0) = 0, f"(1) < 0, f(2) = 5, f'(2) = 0, f"(3) > 0, f(4) = 7 (c) f(0) = 0, f'(0) = -1, f'(1) = 0, f(3) = 0, f'(3) = 2, f"(3) = 0, f"(4) < 0 (d) f(0) = 1, f'(0) = 1, f''(0) = 1 and f"(x) increases as x increases (e) f(0) = 1, f'(0) = 0, f'(x) < 0 for 0 < x < 5, f(5) = f'(5) = 0 (f) f(0) = 3, f'(0) = -2, f"(x) > 0 for 0 < x < 5, f(5) = f'(5) = 0

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Before sketching a possible graphs with given conditions about derivative requirements, we need to know that:

If f(x)=0f'\left(x \right)=0, then f(x)f\left(x \right) has maximum or minimum and the slope is m=0m=0.

If f(x)<0f'\left(x \right)<0, then f(x)f\left(x \right) is decreasing.

If f(x)>0f'\left(x \right)>0, then f(x)f\left(x \right) is increasing.

If f(x)<0f''\left(x \right)<0, then f(x)f\left(x \right) is concave down (\cap).

If f(x)>0f''\left(x \right)>0, then f(x)f\left(x \right) is concave up (\cup).

If f(x)=0f''\left(x \right)=0, then xx is the inflection point, i.e. function changes from concave down to concave up and vice-versa.

Further, if there is given condition as f(x)=yf\left(x \right)=y, then we will draw the point (x,y)\left(x,y\right) and through that point function must pass.

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