E1.

Find the derivative function \(f'(x)\) of \(f(x)\) and evaluate the derivative at point \(x = 1\).

\[f(x) = x^2 + 5\]

Solution:

\[f'(x) = 2x, \qquad f'(1) = 2\]

E2.

Find the derivative function \(f'(x)\) of \(f(x)\) and evaluate the derivative at point \(x = 1\).

\[f(x) = 2x^3 - 6x^2 + 3\]

Solution:

\[f'(x) = 6x^2 - 12x, \qquad f'(1) = 6 - 12 = -6\]

E3.

\[f(x) = 2x^{-3} + \frac{x^2}{2}\]

Solution:

\[f'(x) = -6x^{-4} + x\]

E4.

\[f(x) = \sqrt{3x} + 4 \sqrt{x^3} + \frac{1}{\sqrt{x}}\]

Solution:

\[f'(x) = \frac{\sqrt{3}}{2 \sqrt{x}} + 6\sqrt{x} - \frac{1}{2\sqrt{x}^3}\]

E5.

\[f(x) = \frac{e^x}{2} + 3\log x\]

Solution:

\[f'(x) = \frac{e^x}{2} + \frac{3}{x}\]

E6.

\[f(x) = e^{x/2} + \log x^2\]

Solution:

\[f'(x) = \frac{e^{x/2}}{2} + \frac{2x}{x^2}\]

E7.

\[f(x) = (x^2 + 2)(3x^3 - 5x)\]

Solution:

\[f(x) = h(x) g(x), \quad h(x) = x^2 + 2, \quad g(x) = 3x^3 - 5x\] \[h'(x) = 2x, \quad g'(x) = 9x^2 - 5\]

Use the product rule \(f'(x) = h'(x) g(x) + h(x) g'(x)\).

\[f'(x) = 2x(3x^3 - 5x) + (x^2 + 2)(9x^2 - 5) = 6x^4 - 10x^2 + 9x^4 - 5x^2 + 18x^2 - 10 = 15x^4+3x^2 - 10 \]

E8.

Does the function from E1 have a local minima or maxima? Where? Is any of these global?

Solution:

Find critical points: \(f'(x) = 0\)

\[f'(x) = 2x = 0 \quad \Rightarrow \quad x=0\] What kind of critical point is it?

Option 1: look at function evaluations: \[f(0) = 5, \quad f(1) = 6, \quad f(-1) = 6 \quad \Rightarrow \quad x=0 \text{ is a local minimum}\] Option 2: look at 2nd order derivative: \[f''(x) = 2, \qquad f''(0) = 2 > 0, \qquad \Rightarrow \qquad x=0 \text{ is a local minimum}\] Moreover, the point \(x=0\) is a local and global minimum because the function \(f\) is convex over the whole domain. This is because the 2nd derivative is always bigger than zero \(f''(x) = 2 > 0, \forall x\)

E9.

Does the function from E2 have a local minima or maxima? Where? Is any of these global?

Solution:

Find critical points: \(f'(x) = 0\)

\[f'(x) = 6x^2 - 12x = 6x(x-2) = 0 \quad \Rightarrow \quad x=0 \text{ or } x=2\] What kind of critical point is it?

Option 1: look at function evaluations: \[f(0) = 3, \quad f(-1) = -5 \quad f(1) = -1, \quad \Rightarrow \quad x=0 \text{ is a local maximum}\] \[f(2) = -5, \quad f(1) = -1, \quad f(3) = 3 \quad \Rightarrow \quad x=2 \text{ is a local minimum}\]

Option 2: look at 2nd order derivative: \[f''(x) = 12x - 12, \qquad f''(0) = -12 < 0, \qquad \Rightarrow \qquad x=0 \text{ is local a maximum}\] \[f''(2) = 12 > 0, \qquad \Rightarrow \qquad x=2 \text{ is local a minimum}\]