Let f(x) = x⁴ – 5x² + 4

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

To find critical points, $f^{\prime }\left(x\right)=0$

⇒ 4x³ – 10x = 0

$\Rightarrow x=0,\pm \frac{\sqrt{1}0}{2}$

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

$f\left(x\right)=f\left(a\right)+f^{\prime }\left(a\right)\left(x-a\right)+\frac{f^{\prime }\left(a\right)}{2}{\left(x-a\right)}^2+\frac{f^{‴}\left(a\right)}{3!}{\left(x-a\right)}^3+\dots$

Calculation:

Here a = 3

$f\left(x\right)=f\left(3\right)+f^{\prime }\left(3\right)\left(x-3\right)+\frac{f^{\prime }\left(3\right)}{2}{\left(x-3\right)}^2+\frac{f^{‴}\left(3\right)}{3!}{\left(x-3\right)}^3+\dots$

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)

The conditions to apply mean value theorem are

(i) f(x) is continuous

(ii) f(x) is differentiable

Given function, f(x) = |x|

f(x) is continuous

f(x) is not differentiable at x = 0.

Therefore, mean value theorem is not applicable.

f(x) = e^-x sin(100x)

From the above graph f(x) has a minimum value at $\mathrm{x}=\frac{3\pi }{200}$

$\therefore \mathrm{f}\left(\mathrm{x}=\frac{3\pi }{200}\right)={\mathrm{e}}^{\frac{-3\pi }{200}}\sin \left(\frac{3\pi }{2}\right)$

$\therefore f(x=\frac{3\pi }{200})=-0.9539$

∴ The minimum value of f(x) is 0.9539

Alternate solution:

f(x) = e-x sin(100x)

f'(x) = -e^-x sin(100x) + (e^-xcos (100x) × 100)

for minima f'(x)

-e-x sin(100x) + (e-xcos (100x) × 100) = 0

e-x sin(100x) = e-xcos (100x) × 100

tan (100x) = 100

$x=\frac{1}{100}\times {\tan }^{-1}(100)$

x = 0.0156

Minimum value is 

f(0.0156) = e^-0.0156 sin(100 × 0.0156) = 0.9844

Note: Different values are coming as maxima and minima method gives approximate results.

The greatest amount of light may be admitted means that the area of the semi-circle so that one side of the rectangle is 2x ft. (as shown in figure). Let the other side of the rectangle by y ft. Then the perimeter of the whole figure = πx + 2x + 2y = 40

$\begin{array}{l}area\left(A\right)=\frac{1}{2}\pi {\mathrm{x}}^2+2xy\\ \Rightarrow \mathrm{A}=\frac{1}{2}\pi {\mathrm{x}}^2+\mathrm{x}\left[40-\pi \mathrm{x}-2\mathrm{x}\right]\\ =40\mathrm{x}-\left(\frac{\pi }{2}+2\right){\mathrm{x}}^2\end{array}$  

Then $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}=40-\left(\pi +4\right)\mathrm{x}$

For A to be maximum or minimum, we must have

 $\frac{\mathrm{d}\mathrm{A}}{\mathrm{d}\mathrm{x}}=0$

⇒ 40 – (π + 4)x = 0

$\begin{array}{l}\Rightarrow \mathrm{x}=\frac{40}{\left(\pi +4\right)}\\ \Rightarrow \mathrm{y}=\frac{1}{2}\left[40-\left(\pi +2\right)\mathrm{x}\right]\\ =\frac{1}{2}\left[40-\left(\pi +2\right)\frac{40}{\left(\pi +4\right)}\right]=\frac{40}{\pi +4}\end{array}$  

⇒ x = y

$\frac{{\mathrm{d}}^2\mathrm{A}}{\mathrm{d}{\mathrm{x}}^2}=\left(\pi +4\right)$ which is negative

Concept:

Using Lagrange's Mean Value Theorem:

$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

$f^{\prime }\left(x\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$

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

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

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

$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$     ---(1)

Now,

f(a) = f(0) = 0

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

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

f'(x) = 3x² – 6x + 2

f’(c) = 3c² – 6c + 2

Put in equation (1)

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

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

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].

$f^{\prime }\left(x\right)=2\left(x-3\right)$, 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:

$f^{\prime }\left(c\right)=\frac{f\left(4\right)-f\left(3\right)}{4-3}=1$

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

$f^{\prime }\left(x\right)=2\left(x-3\right)$ = 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 $\left(\frac{7}{2},\frac{1}{4}\right)$