(1) The domain of f (x) is R \{−1,1}.
(2) Since f (−x,−y) = f (x, y) , the curve is symmetric about the origin.
(3) Putting y = 0, we get x = 0 . Hence the x -intercept is (0, 0).
(4) Putting x = 0, we get y = 0. Hence the y -intercept is (0, 0).
(5) To determine monotonicity, we find the first derivative as \(f'(x)\frac { -3\left( { x }^{ 2 }+1 \right) }{ { ({ x }^{ 2 }-1) }^{ 2 } } \)
Hence, f'( x) does not exist at x = −1,1. Therefore, critical numbers are x = −1, 1. The intervals of monotonicity is tabulated

6) Since there is no sign change in f'( x) when passing through critical numbers. There is no local extrema.
(7) To determine the concavity, we find the second derivative as \(f"(x)=\frac { 6x({ x }^{ 2 }-3) }{ { ({ x }^{ 2 }-1) }^{ 3 } } \) f"( x) = 0\(\Rightarrow\) x = 0 and f"(x) does not exist at x = −1, 1.
The intervals of concavity is tabulated.

(8) As x = −1 and 1are not in the domain of f(x) and at x = 0, the second derivative is zero and f"(x) changes its sign from positive to negative when passing through x = 0. Therefore, the point of inflection is (0, f (0)) = (0,0) .

(9) \(\underset { x\rightarrow \pm \infty }{ lim } f(x)=\underset { x\rightarrow \pm }{ lim } \cfrac { 3x }{ { x }^{ 2 }-1 } =\underset { x\rightarrow \pm \infty }{ lim } \cfrac { 3 }{ x\frac { 1 }{ x } } =0\) Therefore y = 0 is a horizontal asymptote.
Since the denominator is zero, when x = 士1.\(\underset { x{ \rightarrow -1 }^{ - } }{ lim } \frac { 3x }{ { x }^{ 2 }-1 } -\infty ,\underset { x{ \rightarrow -1 }^{ + } }{ lim } \frac { 3x }{ { x }^{ 2 }-1 } =\infty ,\underset { x{ \rightarrow 1 }^{ - } }{ lim } \frac { 3x }{ { x }^{ 2 }-1 } -\infty ,\underset { x{ \rightarrow 1 }^{ + } }{ lim } \frac { 3x }{ { x }^{ 2 }-1 } =\infty .\)
Therefore x = −1 and x = 1 are vertical asymptotes.
The rough sketch of the curve is shown on the right side.

Factorising the given function, we have
y = f (x) = (x − 3)(x² + 3x + 3).
(1) The domain and the range of the given function f(x) are the entire real line.
(2) Putting y = 0, we get the x = 3. The other two roots are imaginary. Therefore, the x -intercept is (3, 0) . Putting x = 0, we get y = −9. Therefore, the y-intercept is (0, −9)
(3) f'(x) = 3(x² -2) and hence the critical points of the curve occur at x = \(\pm \sqrt { 2 } \)
(4) f"(x) = 6x. Therefore at x =\(\sqrt{2}\) the curve has a local minimum because f"(\(\sqrt{2}\)) = 6\(\sqrt{2}\) > 0. Then local minimum is f(\(\sqrt{2}\)) = -4\(\sqrt{2}\)-9.
Similarly x = -\(\sqrt{2}\) the curve has a local maximum because f"(-\(\sqrt{2}\)) = -6\(\sqrt{2}\) < 0. The local maximum is f (-\(\sqrt{2}\)) = 4\(\sqrt{2}\) - 9.
(5) Since f "(x) = 6x > 0, ∀x > 0 the function is concave upward in the positive real line. As f"(x) = 6x < 0,∀x < 0 the function is concave downward in the negative real line.
(6) Since f"(x) = 0 at x = 0 and f′′(x) changes its sign when passing through x = 0. Therefore the point of inflection is (0, f (0)) = (0, 9).
(7) The curve has no asymptotes.
The rough sketch of the curve is shown on the right side.

Let the given function be written as f (x) = x² ( 10− x)² . Now
f(x) = x²(100 - 20x + x²) = x⁴-20x³+100x²
Therefore, f'(x) = 4x³- 60x³ + 200x = 4x(x²-15x+50)
f'(x) = 4x(x²-15x + 50) = 0 ⇒ x = 0, 5, 10
and f"(x) = 12x²-120x + 200
The stationary points of f(x) are x = 0, 5, 10 at these points the values of f′′(x) are respectively 200, −100 and 200. At x = 0, it has local minimum and its value is f(0) = 0. At x = 5, it has local maximum and its value is f(5) = 625. At x = 10, it has local minimum and its value is f(10) = 0.

We have,
f'(x) = 4x³+32 = 0 gives x³ = -8
⇒ x = −2
and f′′(x) = 12 x² 
As f''(−2)>0, the function has local minimum at x = −2. The local minimum value is f (−2) = −48 
Therefore, the extreme point is (−2, −48) .

We have, \(P=\frac{234+16x}{x+3}\)
Therefore, \(\frac{dP}{dt}= - \frac{186}{(x+3)^{2}}\times \frac{dx}{dt}\).
Substituting \(x=90, \frac{dx}{dt}=15\) we get\(\frac{dP}{dt}= -\frac{186}{93^{2}}\times 15= -\frac{10}{31}\approx -0.32\) repee/ week.
That is the price is changing, in fact decreasing at the rate of Rs. 0.32 per unit.

Given that s(t) = t³ − 6t² + 9t + 1. On differentiating, we get v(t) = 3t² -12t + 9 and a(t) = 6t −12.
(i) The particle is at rest when v(t) = 0 . Therefore, v(t) = 3(t −1)(t − 3) = 0 gives t = 1 and t = 3.
(ii) The particle changes direction when v (t) changes its sign. Now.
if 0 ≤ t < 1 then both (t −1) and (t − 3) < 0 and hence, v(t) > 0.
If 1< t < 3 then (t −1) > 0 and (t − 3) < 0 and hence, v(t) < 0.
If t > 3 then both (t −1) and (t − 3) > 0 and hence, v(t) > 0.
Therefore, the particle changes direction when t = 1 and t = 3.
(iii) The total distance travelled by the particle from time t = 0 to t = 2 is given by,
|s(0) − s(1)| + |s(1) − s(2)| = |1− 5 | + | 5 − 3| = 6 metres.

The average rate of change in an interval [a, b] is \(\frac { f(b)-f(a) }{ b-a } \) whereas, the instantaneous rate of change at a point x is f′(x) for the given function. They are respectively, b + a and 2x.

We have,
\(\underset { x\rightarrow { 0 }^{ - } }{ lim } =-\infty \ and\ \underset { x\rightarrow { 0 }^{ x } }{ lim } =\frac { 1 }{ x } =\infty \). Hence, the required vertical asymptote is x = 0 or the y -axis.
As the curve is symmetric with respect to both the axes, y = 0 or the x -axis is also an asymptote.
Hence this (rectangular hyperbola) curve has both the vertical and horizontal asymptotes.

If we directly substitute x = 0 we get an indeterminate form \(\frac{0}{0}\) and hence we apply the l’Hôpital’s rule to evaluate the limit as
\(\underset{x\rightarrow 0}{lim}(\frac{sin \ mx}{x})\)=\(\underset{x\rightarrow 0}{lim}(\frac{m\times cos \ mx}{1})\)
= m
The next example tells that the limit does not exist.

If we put directly x = a we observe that the given function is in an indeterminate form \(\frac00\). 
As the numerator and the denominator functions are polynomials they both are differentiable.
Hence by an application of the l’Hôpital Rule we get,
\(\underset{x\rightarrow a}{lim}(\frac{x^{n}-a^{n}}{x-a})\) = \(\underset{x\rightarrow a}{lim}(\frac{n\times x^{n-1}}{1})\)
= \(n \times a^{n-1} \).

We are given that, T = 10x − x²
Hence, the rate of change at any distance from one end is given by \(\frac{dT}{dx}=10-2x \) 
The mid point of the rod is at x = 5
Substituting x = 5, we get \(\frac{dT}{dx}=0\)

Let f (x) = sin x which is a differentiable function in any open interval. Consider an interval \([\alpha, \beta]\). Applying the mean value theorem there exists  \(c \in (\alpha, \beta)\) such that, 
\(\frac{sin \beta - sin \alpha}{\beta-\alpha}=f'(c)=cos(c)\)
Therefore, \(\frac{sin \beta - sin \alpha}{\beta-\alpha}=|cos(c)|\le1\)
Hence, \(|sin\alpha-sin\beta|\le |\alpha-\beta|\)
Remark
If we take \(\beta=0\) in the above problem, we ge \(|sin \alpha|\le |\alpha|\) 

P(x) = \(x^{4}-6x^{3}-11x^{2}+24x+28\), \(\alpha\) = 2, \(\beta\) = 7 
and observing \(\frac{P'(x)}{2}=2x^{3}-9x^{2}-11x+12=Qx\), (say).
This implies that there is a zero of the polynomial Q(x) in the interval (2, 7)
For verification,
Q(2) = 16 - 36 - 22 +12 = 28 - 58 = -30 < 0
Q(7) = 686 - 441 - 77 +12 = 698 - 518 = 180 > 0
From this we may see that there is a zero of the polynomial Q(x) in the interval (2, 7)

Let f (x) = x⁴ + 2x³-2
Then f (x) is continuous in [0, 1] and differentiable in (0, 1)  
Now, f'(x) = 4x³+6x² 
If f'(x) = 0, then
2x²(2x+3) = 0
Therefore,  \(x=0, -\frac{3}{2}\) but  \(0,-\frac{3}{2} \notin (0,1)\).
Thus, \(f'(x)>0, \forall x\in (0,1)\).
Hence by the Rolle’s theorem there do not exist \(a,b \in(0,1)\) such that, f(a) = 0 = f(b). Therefore the equation f(x ) = 0 cannot have two roots in the interval (0, 1) . But, f (0, 2) = −2 < 0 and f (1) = 1 > 0 tells us the curve y f = (x) crosses the x -axis between 0 and 1 only once by the Intermediate value theorem. Therefore the equation x⁴ + 2x³ − 2 = 0 has only one real root in the interval (0, 1) .

Observe that, f(0) = 0 = f (1), is continuous in the interval [0,1] and is differentiable in (0,1). Now,
 \(f'{x}=2x(1-x)(1-2x)\).
Therefore,  \(f'(c)=0 \) gives c = 0, 1 and \(\frac{1}{2}\)
which \(\Rightarrow c= \frac{1}{2}\in (0,1)\).