I.Let f(x)= ) 1+4x-x^2&forxleqslant 3 (x+5)/(2)&forxgt 3 (i) Is f a continuous function on the set of real numbers R? (ii) Is f a differentiable function on R? (iii) Find the maximum value and minimum values of f(x) for xin [0,4] (iv) Plot the graph of f(x) on [-4,4]

(i) To determine if the function $f(x)$ is continuous on the set of real numbers $\mathbb{R}$, we need to check if the function is continuous at every point in $\mathbb{R}$.<br /><br />The function $f(x)$ is defined as a piecewise function with two different expressions for different intervals of $x$. To check the continuity, we need to ensure that the two expressions are equal at the boundary point $x = 3$.<br /><br />For $x \leq 3$, the function is defined as $f(x) = 1 + 4x - x^2$.<br /><br />For $x > 3$, the function is defined as $f(x) = \frac{x+5}{2}$.<br /><br />To check the continuity at $x = 3$, we need to evaluate the left-hand limit and the right-hand limit as $x$ approaches 3.<br /><br />Left-hand limit: $\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (1 + 4x - x^2) = 1 + 4(3) - 3^2 = 4$<br /><br />Right-hand limit: $\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} \frac{x+5}{2} = \frac{3+5}{2} = 4$<br /><br />Since the left-hand limit and the right-hand limit are equal, the function $f(x)$ is continuous at $x = 3$.<br /><br />Therefore, the function $f(x)$ is continuous on the set of real numbers $\mathbb{R}$.<br /><br />(ii) To determine if the function $f(x)$ is differentiable on $\mathbb{R}$, we need to check if the function is differentiable at every point in $\mathbb{R}$.<br /><br />The function $f(x)$ is defined as a piecewise function with two different expressions for different intervals of $x$. To check the differentiability, we need to ensure that the two expressions have the same derivative at the boundary point $x = 3$.<br /><br />For $x \leq 3$, the function is defined as $f(x) = 1 + 4x - x^2$.<br /><br />For $x > 3$, the function is defined as $f(x) = \frac{x+5}{2}$.<br /><br />To check the differentiability at $x = 3$, we need to evaluate the left-hand derivative and the right-hand derivative as $x$ approaches 3.<br /><br />Left-hand derivative: $\frac{d}{dx} (1 + 4x - x^2) = 4 - 2x$<br /><br />Right-hand derivative: $\frac{d}{dx} \frac{x+5}{2} = \frac{1}{2}$<br /><br />Since the left-hand derivative and the right-hand derivative are not equal, the function $f(x)$ is not differentiable at $x = 3$.<br /><br />Therefore, the function $f(x)$ is not differentiable on $\mathbb{R}$.<br /><br />(iii) To find the maximum and minimum values of $f(x)$ for $x \in [0, 4]$, we need to evaluate the function at the critical points and endpoints of the interval.<br /><br />The critical points occur where the derivative of the function is zero or undefined. In this case, the derivative is undefined at $x = 3$, so we need to evaluate the function at $x = 3$.<br /><br />For $x \leq 3$, the function is defined as $f(x) = 1 + 4x - x^2$.<br /><br />For $x > 3$, the function is defined as $f(x) = \frac{x+5}{2}$.<br /><br />Evaluating the function at the critical point $x = 3$:<br /><br />$f(3) = 1 + 4(3) - 3^2 = 4$<br /><br />Now, we need to evaluate the function at the endpoints $x = 0$ and $x = 4$.<br /><br />$f(0) = 1 + 4(0) - 0^2 = 1$<br /><br />$f(4) = \frac{4+5}{2} = \frac{9}{2} = 4.5$<br /><br />Therefore, the maximum value of $f(x)$ for $x \in [0, 4]$ is $4.5$, and the minimum value is $1$.<br /><br />(iv) To plot the graph of $f(x)$ on $[-4, 4]$, we need to evaluate the function for a range of values within the interval $[-4, 4]$.<br /><br />For $x \leq 3$, the function is defined as $f(x) = 1 + 4x - x^2$.<br /><br />For $x the defined as $f(x