Exercises 1

This page allows you to practice some exercises on Differentiation. If you notice that you have difficulties, we advise you to go over the corresponding sections in the book.

These questions correspond to the following sections: - Section 6.1 - Slopes of Curves - Section 6.2 - Tangents and Derivatives - Section 6.3 - Increasing and Decreasing Functions - Section 6.4 - Economic Applications

Question 1

Let $f (x) = 3x^2 + 2x − 1$.

(a) Show that $\frac{f (x + h) − f (x)}{h} = 6x + 2 + 3h$ for $h \neq 0$, and use this result to find $f'(x)$.

(b) Find in particular $f'(0)$, $f'(-2)$, and $f'(3)$. Find also the equation of the tangent to the graph at the point $(0, -1)$.

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(a) Note that $f(x+h) - f(x) = [3(x+h)^2 + 2(x+h) -1] - [3x^2 + 2x -1] = [3(x^2 +2hx + h^2) + 2x + 2h - 1] -[3x^2 + 2x -1] = 6hx + 3h^2 +2h.$

Now, we obtain $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} (6x + 3h + 2) = 6x + 2$.

(b) $f'(0) = 2, f'(-2) = -10, f'(3) = 20$. The tangent equation is $y =2x - 1$.

Question 2

For $f (x) = 1/x$, show that $\frac{f (x + h) − f (x)}{h} = -\frac{1}{x(x+h)}$ and use this to show that$f (x) = x^{-1} \implies f'(x) = -x^{-2}$.

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$\frac{f(x+h) - f(x)}{h} = \frac{1/(x+h) - 1/x}{h} = \frac{x - (x+h)}{hx(x+h)} = \frac{-h}{hx(x+h)} = -\frac{1}{x(x+h)} \underset{h \to 0}{\rightarrow} -\frac{1}{x^2} = f' (x).$

Question 3

In each case below, find the slope of the tangent to the graph of $f$ at the specified point:

(a) $f (x) = x^2 − 1,$ at $(1, 0)$

(b) $f (x) = x^3 − 2x,$ at $(0, 0)$

(c) $f (x) = x +\frac{1}{x},$ at $(−1, −2)$

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(a) $f'(1) = 2 \cdot 1 = 2$.

(b) $f'(0) = 3 \cdot 0^2 - 2 = -2$.

(x) $f' (-1) = 1 - [1/(-1)^{-2}] = 1 -1 = 0$.

Question 4

Let $f(x) = \sqrt{x} = x^{1/2}$.

(a) Show that $(\sqrt{x + h} - \sqrt{x})(\sqrt{x + h} + \sqrt{x}) = h$.

(b) Use the result in part (a) to show that the Newton quotient of $f(x)$ is $1/(\sqrt{x + h} + \sqrt{x})$.

(c) Use the result in part (b) to show for $x > 0$ one has $f'(x) = \frac{1}{2\sqrt{x}} = \frac{1}{2}x^{-1/2}$.

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(a) $x+ h + \sqrt{x}\sqrt{x+h} - \sqrt{x}\sqrt{x+h} - x = h$. Simply expand the left-hand side.

(b) By rearranging the identity in (a), we obtain:

$\frac{f(x+h) - f(x)}{h} = \frac{\sqrt{x+h} - \sqrt{x}}{h} = 1/(\sqrt{x+h} + \sqrt{x})$.

(c) Note that $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \frac{1}{2\sqrt{x}} = \frac{1}{2}x^{-1/2}$ .

Question 5

(a) Find the values of $x$ at which $f(x) = x^2 - 4x + 3$ is increasing/decreasing.

(b) Examine where $f(x) = -x^3 + 4x^2 - x - 6$ is increasing/decreasing.

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(a) $f'(x) = 2x - 4$, so the function $f(x)$ is decreasing in $(-\infty, 2]$ and increasing in $[2, \infty)$.

(b) $f'(x) = -3x^2 8x -1 = −3(x − x_0)(x − x_1 )$, where $x_0 = \frac{1}{3}(4 - \sqrt{13}) \approx 0.13$ and $x_{1} = \frac{1}{3}(4+\sqrt{13}) \approx 2.54.$ Then $f(x)$ is decreasing in $(-\infty, x_{0}]$, increasing in $[x_{0}, x_{1}]$, and decreasing in $[x_{1}, \infty)$.