Applications of Derivatives practice questions

Question 1

The function $f(x) = x^2 - 4x + 5$ has its minimum at:

Accepted Answer

$f'(x) = 2x - 4 = 0 \Rightarrow x = 2$. Since $f''(x) = 2 > 0$, this is a minimum.

Answer

$x = -2$

Answer

$x = 2$

Answer

$x = 4$

Answer

$x = 5$

Question 2

The function $f(x) = x^2 + 2x$ is increasing on:

Accepted Answer

$f'(x) = 2x + 2 > 0 \Leftrightarrow x > -1$. So $f$ is increasing on $(-1, +\infty)$.

Answer

$\mathbb{R}$

Answer

$(-\infty, -1)$

Answer

$(-1, +\infty)$

Answer

$(0, +\infty)$

Question 3

The slope of the tangent to $f(x) = x^2$ at $x = 3$ is:

Accepted Answer

$f'(x) = 2x$, so $f'(3) = 6$.

Answer

$0$

Answer

$3$

Answer

$6$

Answer

$9$

Question 4

The function $f(x) = x^{2}$ is:

Accepted Answer

$f'(x) = 2x$ and $f''(x) = 2 > 0$ for all $x \in \mathbb{R}$.

Since $f''(x) > 0$ everywhere, $f$ is convex (concave up) on all of $\mathbb{R}$ — no inflection point.

Answer

convex on $\mathbb{R}$ ($f''(x) > 0$)

Answer

concave on $\mathbb{R}$

Answer

convex only on $(0,\infty)$

Answer

has an inflection point at $x = 0$

Question 5

The function $f(x) = x^{3} - 3x$ is **decreasing** on the interval:

Accepted Answer

$f'(x) = 3x^{2} - 3 = 3(x - 1)(x + 1)$.

$f'(x) < 0$ when $(x-1)(x+1) < 0$, i.e. on $(-1, 1)$.

So $f$ is decreasing on $(-1, 1)$.

Answer

$(-\infty, -1)$

Answer

$(-1, 1)$

Answer

$(1, \infty)$

Answer

$\mathbb{R}$

Question 6

For the same position $x(t) = t^3 - 6t^2 + 9t$, the acceleration at $t = 1$ is:

Accepted Answer

$a(t) = x''(t) = 6t - 12$. At $t = 1$: $a = -6$ m/s².

Answer

$-6$ m/s²

Answer

$0$ m/s²

Answer

$6$ m/s²

Answer

$9$ m/s²

Question 7

The function $f(x) = x^3$ is concave on:

Accepted Answer

$f''(x) = 6x$. $f'' < 0 \Leftrightarrow x < 0$. So $f$ is concave on $(-\infty, 0)$.

Answer

$\mathbb{R}$

Answer

$(-\infty, 0)$

Answer

$(0, +\infty)$

Answer

nowhere

Question 8

The point of inflection of $f(x) = x^3 - 3x$ is at:

Accepted Answer

$f'(x) = 3x^2 - 3$, $f''(x) = 6x$. $f''$ changes sign at $x = 0$, so the inflection is at $x = 0$.

Answer

$x = -1$

Answer

$x = 0$

Answer

$x = 1$

Answer

$f$ has no inflection

Question 9

The function $f(x) = -x^2 + 4x - 1$ has a local maximum at:

Accepted Answer

$f'(x) = -2x + 4 = 0 \Rightarrow x = 2$. $f''(x) = -2 < 0$, confirming a local maximum.

Answer

$x = -2$

Answer

$x = 0$

Answer

$x = 2$

Answer

$x = 4$

Question 10

A particle's position is $x(t) = t^3 - 6t^2 + 9t$ (in metres, $t$ in seconds). The velocity at $t = 2$ is:

Accepted Answer

$v(t) = 3t^2 - 12t + 9$. At $t = 2$: $v = 12 - 24 + 9 = -3$ m/s (the particle is moving backward).

Answer

$-3$ m/s

Answer

$0$ m/s

Answer

$3$ m/s

Answer

$9$ m/s

Applications of Derivatives

Problems & worked solutions

Problem 1 — Applications of Derivatives

Problem 2 — Applications of Derivatives

Problem 3 — Applications of Derivatives

Problem 4 — Convexity & Concavity (2nd Derivative Table)

Problem 5 — Monotonicity (1st Derivative Table)

Problem 6 — Applications of Derivatives

Problem 7 — Applications of Derivatives

Problem 8 — Applications of Derivatives

Problem 9 — Applications of Derivatives

Problem 10 — Applications of Derivatives

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