📝 Graphing functions using calculus (22 MCQs)

📖 Explanation: Using the chain rule, differentiate the exponent $-x^{2}/2$ to get $-x$. Multiply this factor by the original function $e^{-x^{2}/2}$. Hence the derivative is $-x e^{-x^{2}/2}$, which matches option A. The other options miss the sign or the factor from the chain rule.

📖 Explanation: Setting $x = 0$ gives $y = e^{-(0)^{2}/2}=e^{0}=1$. Therefore the point where the graph meets the y‑axis is $(0,1)$. The other listed values do not result from substituting $x = 0$ into the function.

📖 Explanation: Replacing $x$ by $-x$ leaves the expression unchanged because $(-x)^{2}=x^{2}$. Hence the function satisfies $f(-x)=f(x)$ and is symmetric with respect to the y‑axis, i.e., it has even symmetry. The other choices do not describe this behavior.

📖 Explanation: For $x>0$, the factor $-x$ is negative while $e^{-x^{2}/2}$ is always positive. Their product is therefore negative, meaning f&#039;(x)<0. A negative derivative indicates the function is decreasing on any interval where $x>0$.}

📖 Explanation: Concavity changes where the second derivative changes sign, i.e., where $(x^{2}-1)=0$. Solving gives $x^{2}=1$ so $x = \pm1$. At these points the factor $e^{-x^{2}/2}$ is positive, so the sign of the second derivative is determined solely by $x^{2}-1$.}

📖 Explanation: An x‑intercept requires the function value to be zero. Since the exponential expression is always positive, it never reaches zero, so the graph never crosses the x‑axis. Hence the function has no x‑intercepts.

📖 Explanation: For large $|x|$, the exponent $-x^{2}/2$ grows in magnitude quadratically, while $-x$ grows linearly. A larger negative exponent forces the exponential term toward zero more rapidly. Consequently, $e^{-x^{2}/2}$ approaches zero faster than $e^{-x}$.}

📖 Explanation: Reflecting across the y‑axis replaces $x$ with $-x$. Since $(-x)^{2}=x^{2}$, the expression inside the exponential stays the same, yielding the original function. Thus the reflected graph is identical to the original, confirming option D.

📖 Explanation: A horizontal asymptote at $y=0$ means the function values approach zero as $x$ goes to positive or negative infinity. The correct limit statement is $\lim_{x\to\infty} e^{-x^{2}/2}=0$ (and similarly for $-\infty$). The other limits are incorrect for this asymptote.

📖 Explanation: At a relative maximum, the derivative changes from positive (function increasing) to negative (function decreasing). Therefore, just left of $x=0$ the derivative is positive, and just right it is negative, matching option C.

📖 Explanation: Shifting left replaces $x$ with $x+1$. The original inflection points at $x=\pm1$ become $x+1=\pm1$ → $x=-2$ and $x=0$. Their y‑coordinates are $e^{-(-2)^{2}/2}=e^{-2}$ and $e^{-0}=1$. Hence option A is correct.

📖 Explanation: Even symmetry requires $f(-x)=f(x)$. Among the choices, $\cos x$ satisfies this property because cosine is an even function. The others are odd or not symmetric. Therefore, option C matches the symmetry of the given exponential function.

📖 Explanation: A critical point occurs where the derivative is zero. For the exponential, f&#039;(x)=-x e^{-x^{2}/2}=0 gives only $x=0$. A cubic with positive leading coefficient generally has two critical points (a local max and min). Hence the exponential has fewer critical points than the cubic.

📖 Explanation: Increasing the denominator reduces the magnitude of the exponent for a given $x$, causing the function to decay more slowly. This spreads the bell shape, making it wider. Hence the graph broadens, matching option B.

📖 Explanation: The second derivative of the exponential is $(x^{2}-1)e^{-x^{2}/2}$, changing sign at $x=\pm1$. The logarithm $\ln(x+2)$ has second derivative $-1/(x+2)^{2}$, which is always negative on its domain $(-2,\infty)$. Thus the exponential has changing concavity, whereas the logarithm remains concave down throughout. Option C captures this.

📖 Explanation: At a local maximum, the function is concave down, implying a negative second derivative. For the given function, f&#039;&#039;(0) = (0^{2}-1)e^{0} = -1, confirming the second derivative is negative at the maximum point.

📖 Explanation: The Gaussian integral $\int_{-\infty}^{\infty} e^{-x^{2}/2}\,dx$ equals $\sqrt{2\pi}$ when the exponent is $-x^{2}/2$. However, the standard result $\int_{-\infty}^{\infty} e^{-x^{2}}dx = \sqrt{\pi}$. Scaling the variable by $\sqrt{2}$ yields $\sqrt{2\pi}$. Since the exponent already contains the factor $1/2$, the correct value is $\sqrt{2\pi}$. (Option B was intended as a distractor; the accurate answer is $\sqrt{2\pi}$.)}

📖 Explanation: A vertical asymptote occurs when the function grows without bound as it approaches a finite $x$-value. The exponential function is continuous and finite for every real $x$; it never tends to $\pm\infty$. Hence no vertical asymptote can exist, making option A correct.

📖 Explanation: The derivative of an even function is odd because differentiating the symmetry condition $f(-x)=f(x)$ yields -f&#039;(-x)=f&#039;(x), implying f&#039;(-x)=-f&#039;(x). Therefore, the derivative changes sign with $x$ and is an odd function.

📖 Explanation: The inner function $e^{-x^{2}/2}$ is continuous and differentiable everywhere. The outer sine function is also continuous and differentiable for all real inputs. Composition of continuous (and differentiable) functions preserves those properties, so $g(x)$ is both continuous and differentiable for every real $x$.

📖 Explanation: An inflection point occurs where the second derivative changes sign, indicating a switch in concavity. At $(-1, e^{-1/2})$ the graph moves from concave up to concave down, and at $(1, e^{-1/2})$ it switches back. Thus these points mark curvature transitions, matching option A.

📖 Explanation: The function is even, giving symmetry about the y‑axis, and has a y‑intercept at (0,1). Since the exponential is always positive and tends to zero as $|x|$ increases, the curve forms a smooth bell shape that approaches the x‑axis without crossing it. This description aligns with option A.