15. If f(x)=x^3 , then lim (f(2+h)-f(2))/(h)=underline ( ) A. 4 B. -4 C. 12 D. infty 16. Let f(x)x^3-12x+1 be defined on [0,3] What is the range of f? A. (-infty ,infty ) B. (-infty ,1] C. [-15,1] D. [-15,17) 17. If f(x)=sqrt (2+sqrt (x)) then f^1(x)=underline ( ) A. (1)/(sqrt (2x+sqrt (x)+)) B. (1)/(4sqrt (2x+xsqrt (x))) (x)/(1+x) D. 2x 18. Which of the ff is the set of all critical numbers of f(x)=. f(x)= ) x-1,xlt 1 x^2-x,xgeqslant 1 A. 1 B. (1)/(2) C. (1)/(2),1 D. 19. int _((pi )/(4))^pi /3(1)/(x^2) dx=underline ( ) A. -pi /12 B. -12pi C. -pi D. -1 20. If the sum of n-terms of an arithmetic sequence is 3n^2+5n and a_(n)=164 then what is the nu terms? A. 21 C. 45 B. 27 D. 71 21. What are the respective valued of A ,B and C so that (x-1)/(x^3)+x=(A)/(x)+(Bx+C)/(x^2)+1 A. -1 , 1,1 B. -1 , 1,2 C. 1. -2 o D. 3.1.2 22. Which of the ff is not true about the graph of g(x)=(x^2-1)/(x^2)+1 A. Range of g is (-infty ,1) C. y=1 is H.A B. g is an even function D. Asx-infty ,g(x)-1 23. Let f(x)=x^4-6x^2+1 on [-1,3) The local maximum value of f is __ A f(0) B f(sqrt (3)) C f(3) D f(-sqrt (3))

## Step 1: Evaluating the limit expression<br />### The question asks for the value of $\lim_{h \to 0} \frac{f(2+h) - f(2)}{h}$, where $f(x) = x^3$. We substitute $f(x)$ into the expression: $\lim_{h \to 0} \frac{(2+h)^3 - 2^3}{h}$. Expanding $(2+h)^3$ gives $8 + 12h + 6h^2 + h^3$. Substituting this back into the limit expression, we get $\lim_{h \to 0} \frac{8 + 12h + 6h^2 + h^3 - 8}{h} = \lim_{h \to 0} \frac{12h + 6h^2 + h^3}{h} = \lim_{h \to 0} (12 + 6h + h^2)$. As $h$ approaches 0, the expression approaches 12.<br /><br />## Step 2: Determining the range of f(x)<br />### The function $f(x) = x^3 - 12x + 1$ is defined on the closed interval $[0, 3]$. We check the endpoints and critical points to find the range. $f(0) = 1$. $f(3) = 27 - 36 + 1 = -8$. To find critical points, we take the derivative: $f'(x) = 3x^2 - 12$. Setting $f'(x) = 0$, we get $3x^2 = 12$, so $x^2 = 4$, and $x = \pm 2$. Since we are considering the interval $[0, 3]$, we only consider $x = 2$. $f(2) = 8 - 24 + 1 = -15$. The minimum value is -15 and the maximum is 1. Thus, the range is $[-15, 1]$.<br /><br />## Step 3: Finding the inverse function<br />### Given $f(x) = \sqrt{2 + \sqrt{x}}$, we want to find $f^{-1}(x)$. Let $y = f(x)$, so $y = \sqrt{2 + \sqrt{x}}$. Squaring both sides gives $y^2 = 2 + \sqrt{x}$. Then, $y^2 - 2 = \sqrt{x}$, and squaring again gives $x = (y^2 - 2)^2$. Thus, $f^{-1}(x) = (x^2 - 2)^2$. The question asks for the derivative of the inverse, which isn't provided in the options.<br /><br />## Step 4: Identifying critical numbers<br />### Critical numbers occur where the derivative is zero or undefined. For $x < 1$, $f(x) = x - 1$, so $f'(x) = 1$. For $x \ge 1$, $f(x) = x^2 - x$, so $f'(x) = 2x - 1$. Setting $2x - 1 = 0$, we get $x = \frac{1}{2}$. However, this is not in the domain $x \ge 1$. The derivative is undefined at $x = 1$ due to the piecewise definition. Checking the limit from the left, we have 1, and from the right, we have $2(1)-1=1$. The derivative exists at $x=1$. Thus, there are no critical numbers.<br /><br />## Step 5: Evaluating the definite integral<br />### We evaluate $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{1}{x^2} dx = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} x^{-2} dx = [-x^{-1}]_{\frac{\pi}{4}}^{\frac{\pi}{3}} = -\frac{1}{\frac{\pi}{3}} - (-\frac{1}{\frac{\pi}{4}}) = -\frac{3}{\pi} + \frac{4}{\pi} = \frac{1}{\pi}$.<br /><br />## Step 6: Finding the number of terms<br />### The sum of an arithmetic sequence is given by $S_n = 3n^2 + 5n$. We know that $a_n = 164$. The general term of an arithmetic sequence is $a_n = a_1 + (n-1)d$. The sum can also be expressed as $S_n = \frac{n}{2}(2a_1 + (n-1)d) = \frac{n}{2}(a_1 + a_n)$. We have $S_n = 3n^2 + 5n$. $S_1 = a_1 = 8$. $S_2 = 2a_1 + d = 22$, so $d = 6$. Then $164 = 8 + (n-1)6$, so $156 = 6(n-1)$, and $n-1 = 26$, thus $n = 27$.<br /><br />## Step 7: Partial fraction decomposition<br />### We have $\frac{x-1}{x^3+x} = \frac{x-1}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}$. Multiplying by $x(x^2+1)$, we get $x-1 = A(x^2+1) + (Bx+C)x$. If $x=0$, $-1 = A$. Then $x-1 = -x^2-1 + Bx^2 + Cx$, so $x-1 = (B-1)x^2 + Cx - 1$. Thus, $C=1$ and $B-1=0$, so $B=1$. Therefore, $A=-1$, $B=1$, and $C=1$.<br /><br />## Step 8: Analyzing the graph of g(x)<br />### The function $g(x) = \frac{x^2-1}{x^2+1}$. As $x \to \infty$, $g(x) \to 1$, so $y=1$ is a horizontal asymptote. Since $g(-x) = g(x)$, the function is even. The range is $[-1, 1)$, as the numerator is always less than the denominator.<br /><br />## Step 9: Finding the local maximum<br />### $f(x) = x^4 - 6x^2 + 1$. $f'(x) = 4x^3 - 12x = 4x(x^2 - 3)$. Critical points are $x=0$ and $x=\pm\sqrt{3}$. Since the interval is $[-1, 3)$, we consider $x=0$ and $x=\sqrt{3}$. $f(0) = 1$. $f(\sqrt{3}) = 9 - 18 + 1 = -8$. $f(-1) = 1-6+1 = -4$. $f(3) = 81 - 54 + 1 = 28$. The local maximum on $[-1,3)$ is $f(0)=1$, but the maximum value is $f(3)=28$.