Engineering Mathematics Test 2

Let f(x) = x⁴ – 5x² + 4

\(f'\left( x \right) = 4{x^3} - 10x\)

To find critical points, \(f'\left( x \right) = 0\)

⇒ 4x³ – 10x = 0

\(\Rightarrow x = 0, \pm \frac{{\sqrt 10}}{2}\)

These points are not in the given interval [2, 3].

Thus, f(x) is monotonic in the interval [2, 3]

f(2) = (2)⁴ – 5(2)² + 4 = 0

f(3) = (3)⁴ – 5(3)² + 4 = 40

Minimum value of f(x) = 0

Maximum value of f(x) = 40

Note:

f'(x) is not zero in the given range i.e. in  the interval [2, 3]. So the function is either increasing or decreasing. The given function is increasing in the given range so maximum value will occur at end point i.e. x = 3.

Concept:

Taylor series expansion

\(f\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right) + \frac{{f'\left( a \right)}}{2}{\left( {x - a} \right)^2} + \frac{{f'''\left( a \right)}}{{3!}}{\left( {x - a} \right)^3} + \ldots \)

Calculation:

Here a = 3

\(f\left( x \right) = f\left( 3 \right) + f'\left( 3 \right)\left( {x - 3} \right) + \frac{{f'\left( 3 \right)}}{2}{\left( {x - 3} \right)^2} + \frac{{f'''\left( 3 \right)}}{{3!}}{\left( {x - 3} \right)^3} + \ldots \)

For linear approximation, take only first two terms

f(x) = f(3) + f’(3) (x-3)

f(x) = x³ - 10x²+6, f(3) = -57

f’(x) = 3x² - 20x, f’(3) = -33

∴ f(x) = -57 - 33(x-3) = 42 - 33x = 3(14 - 11x)

Concept:

Using Lagrange's Mean Value Theorem:

\(f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\)

\(f'\left( x \right) = \frac{{f\left( 1 \right) - f\left( 0 \right)}}{{1 - 0}}\)

f(1) = f'(x) + f(0)

\(= \frac{1}{{1 + x}} + 5\)

For maximum value of f(1), x should be minimum.

The minimum value of x interval 0 ≤ x ≤ 1 is 0

\({\left[ {f\left( 1 \right)} \right]_{max}} = \frac{1}{{1 + 0}} + 5 = 6\)

Taylor series expansion

\(\begin{array}{l} f\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \frac{{{f^{\left( n \right)}}\left( a \right)}}{{n!}}{\left( {x - a} \right)^n}\\ f\left( x \right) = f\left( a \right) + {f^I}\left( a \right)\left( {x - a} \right) + \frac{{{f^{II}}\left( a \right)}}{{2!}}{\left( {x - a} \right)^2} + \frac{{{f^{III}}\left( a \right)}}{{3!}}{\left( {x - a} \right)^3} + \ldots \end{array}\)

Here a = 1

f(x) = ln n

\({f^I}\left( x \right) = \frac{1}{x}\)

\({f^{II}}\left( x \right) = - \frac{1}{{{x^2}}}\)

\({f^{III}}\left( x \right) = \frac{2}{{{x^3}}}\)

\({f^{iv}}\left( x \right) = - \frac{6}{{{x^4}}}\)

f^iv (1) = -6

And we have to take the 4^th-degree polynomial i.e n = 4.

∴ The coefficient of (x - 1)⁴ will be:

\(\frac{{{f^4}\left( 1 \right)}}{{4!}} = - \frac{6}{{4!}} = - \frac{1}{4}\)

f(x) = 2x³ – 9x² + 12x + 6

f’(x) = 6x² – 18x + 12 = 0

put f’(x) = 0 to find the point at which maximum and minimum value exists

6x² – 18x + 12 = 0

x² – 3x + 2 = 0

(x – 2)(x – 1 ) = 0

∴ x = 2 or x =1

Interval [2, 3]

f'’(x) = 2x – 3

put x = 2

f’’(x) = 1 > 0 (minimum value might exist)

Also check border values:

Sum of minimum and maximum value = 10 + 15 = 25

For Lagrange’s mean value theorem, there exists at least one real number ‘c’ in (a, b) such that

\(f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\)     ---(1)

Now,

f(a) = f(0) = 0

\(f\left( b \right) = f\left( {\frac{1}{2}} \right) = \frac{1}{2}\left( {\frac{1}{2} - 1} \right)\left( {\frac{1}{2} - 2} \right) = \frac{3}{8}\)

\(f'\left( x \right) = x\left( {{x^2} - 3x + 2} \right) = {x^3} - 3{x^2} + 2x\)

f'(x) = 3x² – 6x + 2

f’(c) = 3c² – 6c + 2

Put in equation (1)

\(3{c^2} - 6c + 2 = \frac{{\frac{3}{8} - 0}}{{\frac{1}{2} - 0}}\)

\(3{c^2} - 6c + 2 = \frac{3}{4}\)

12c² – 24c + 8 = 3

12c² – 24c + 5 = 0

\(c = \frac{{24 \pm \sqrt {{{24}^2} - 12 \times 5 \times 4} }}{{2 \times 12}}\)

c = 1 ± 0.764 = 1.764 or 0.236

But, c = 0.236, since it only lies between 0 and 1/2

f(x) being a polynomial function, it is continuous on [3, 4].

\(f'\left( x \right) = 2\left( {x - 3} \right)\), which exists for all x ∊ [3, 4].

f(x) is differentiable in [3, 4].

We are to find a point on the parabola whose slope equals the slope of the chord/line joining A(3, 0) and B(4, 1).

The slope of the line joining the two points is:

\(f'\left( c \right) = \frac{{f\left( 4 \right) - f\left( 3 \right)}}{{4 - 3}} = 1\)

We have to find the value of x which satisfies the following:

\(f'\left( x \right) = 2\left( {x - 3} \right)\) = 1

Solving the above we find that x = 7/2 is satisfying the above equation:

Solving for y, we get: \(y = {\left( {x - 3} \right)^2} \Rightarrow y = \frac{1}{4}\)

Thus required point is \(\left( {\frac{7}{2},\frac{1}{4}} \right)\)