Unit 2

Calculus -I

Q1) Explain polynominal function.

A1)

A polynomial in the variable x is a function that can be written in the form-

Here are constant.

The term containing the highest power of x are called the leading term

The degree of the polynomial is the power of x in the leading term. degree 0, 1, and 2 are polynomials which are constant, linear and quadratic functions while degree 3, 4, and 5 are polynomials with special names: cubic, quartic, and quintic functions. Polynomials with degree n > 5 are just called degree polynomials.

Q2) Find the domain and range of the function f(x) =

A2)

f(0) = 3/-3  = -1

f(1) = -1

f(2) = -1

f(3) = -1

So that the domain is = {-1, 0, 1, 2, 4,….} and range = {-1, -1, -1, -1,…..}

Q3) find the value of f(2), f(0) and f(3) of the given function-

A3)

Q4) if f(x) = then prove that

A4)

By taking LHS-

Hence proved

Q5) Differentiate the function f(x) = by using the first principal method.

A5)

We know that-

Here

Substituting ( for x gives-

Hence-

Q6) evaluate the

A6) We can simply find the Solutionution as follows,

Q7) evaluate

A7)

Q8) evaluate

A8)

Q9) Find dy/dx of the following functions-

A9)

Let y =

Then-

And

Let y =

Then-

Q10) Differentiate with respect to x.

A10)

Let

Now

Q11) if then find dy/dx.

A11)

Suppose y = u/v where u = x - 1 and v = x + 1 

Then

And

So that-

Q12) if y = then find dy/dx.

A12)

Suppose z =

Now-

So that-

Q13) if y =

A13)

Suppose y = where z =

Q14) If y = log loglog then find dy/dx.

A14)

Suppose y = log u where u = log v and v = log

So that-

Q15) if y =

A15)

Let y = log u where u =

Now

Q16) find the derivative of the function f(x) = .

A16)

Let y = f(x) then

Q17) Find the derivative of

A17)

Let y = then-

Q18) if y = then find

A18)

Here

Difference with respect to x, we get-

Now

Q19) if y = then find .

A19)

Here

y =

Then

Q20) Examine for maximum and minimum for the function f(x) =

A20)

Here the first derivative is-

So that, we get-

Now we will get to know that the function is maximum or minimum at these values of x.

For x = 3

Let us assign to x, the values of 3 – h and 3 + h (here h is very small) and put these values at f(x).

Then-

Which is negative for h is very small

Which is positive

Thus f’(x) changes sign from negative to positive as it passes through x = 3.

So that f(x) is minimum at x = 3 and the minimum value is-

And f(x) is maximum at x = -3.