AP Calculus AB AP Calculus AB Practice Test 2025

Direct substitution gives $\frac{3^2 - 9}{3 - 3} = \frac{0}{0}$, which is indeterminate. We factor the numerator:

For $x \neq 3$, we can cancel the common factor $(x - 3)$:

The correct answer is (b) 6.

We factor both the numerator and the denominator:

• Numerator: $x^2 - 4 = (x - 2)(x + 2)$
• Denominator: $x^2 - x - 2 = (x - 2)(x + 1)$

So $f(x) = \frac{(x - 2)(x + 2)}{(x - 2)(x + 1)}$.

The function is undefined at $x = 2$ and $x = -1$ (where the denominator is zero).

At $x = 2$: both the numerator and denominator are zero, and the common factor $(x - 2)$ can be cancelled. So $\lim_{x \to 2} f(x) = \frac{2 + 2}{2 + 1} = \frac{4}{3}$. Since the limit exists but $f(2)$ is undefined, this is a removable discontinuity.

At $x = -1$: the numerator equals $(-1)^2 - 4 = -3 \neq 0$ while the denominator is zero, so this is a non-removable (vertical asymptote) discontinuity.

The correct answer is (c) $x = 2$.

The limit definition of the derivative at a point $x = a$ is:

This represents the slope of the tangent line to the curve $y = f(x)$ at the point where $x = a$. It is derived from the difference quotient by taking the limit as the interval width $h$ approaches zero.

Option (a) is the difference quotient without the limit. Option (c) is a definite integral, not a derivative. Option (d) has the wrong sign in the numerator, which would give $-f'(a)$ instead of $f'(a)$.

The correct answer is (b).

We apply the power rule $\frac{d}{dx}[x^n] = nx^{n-1}$ to each term:

• $\frac{d}{dx}[3x^5] = 3 \cdot 5x^4 = 15x^4$
• $\frac{d}{dx}[-2x^3] = -2 \cdot 3x^2 = -6x^2$
• $\frac{d}{dx}[7x] = 7$
• $\frac{d}{dx}[-4] = 0$

Therefore: $f'(x) = 15x^4 - 6x^2 + 7$

The correct answer is (a) $15x^4 - 6x^2 + 7$.

We apply the product rule: if $g(x) = u(x) \cdot v(x)$, then $g'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)$.

Here $u(x) = x^2$ and $v(x) = \sin(x)$, so:

• $u'(x) = 2x$
• $v'(x) = \cos(x)$

Applying the product rule:

This can also be written as $2x\sin(x) + x^2\cos(x)$.

The correct answer is (b) $2x\sin(x) + x^2\cos(x)$.

We apply the chain rule. Let $u = 3x^2 + 1$, so $h(x) = u^4$.

By the chain rule: $h'(x) = 4u^3 \cdot \frac{du}{dx}$

We compute: $\frac{du}{dx} = 6x$

Substituting back:

The correct answer is (d) $24x(3x^2 + 1)^3$.

We are given $\frac{dV}{dt} = 12\pi$ cm$^3$/s and need to find $\frac{dr}{dt}$ when $r = 3$ cm.

Starting with the volume formula $V = \frac{4}{3}\pi r^3$, we differentiate both sides with respect to time $t$:

Substituting the known values:

The radius is increasing at a rate of $\frac{1}{3}$ cm/s.

The correct answer is (a) $\frac{1}{3}$ cm/s.

We find the antiderivative of $3x^2 - 2x + 1$:

Now we evaluate using the Fundamental Theorem of Calculus:

The correct answer is (c) 51.

By the Fundamental Theorem of Calculus (Part 1), if $F(x) = \int_a^x f(t)\,dt$, then $F'(x) = f(x)$.

Here, $f(t) = t^3 - 4t$, so:

This is a direct application of the theorem -- the derivative of an integral with a variable upper limit equals the integrand evaluated at that upper limit.

Option (c) would be the second derivative $F''(x) = 3x^2 - 4$. Option (d) is the antiderivative $\frac{x^4}{4} - 2x^2$, which is not what the question asks for.

The correct answer is (a) $x^3 - 4x$.

First, we determine which function is greater on the interval $[0, 2]$. Setting $x^2 = 2x$ gives $x^2 - 2x = 0$, so $x(x - 2) = 0$, meaning $x = 0$ or $x = 2$. For $0 < x < 2$, we check $x = 1$: $y = 1^2 = 1$ and $y = 2(1) = 2$, so $2x > x^2$ on this interval.

The area between the curves is:

The correct answer is (b) $\frac{4}{3}$.