Exercises — Natural exponential function eˣ — graph, derivative preview

L1.1 $e^0=1$. Why: any nonzero base raised to the power $0$ is $1$ — that's what the exponent $0$ means (an empty product). The curve therefore crosses the $y$-axis at height $1$.

L1.2 False. $e^x$ is a positive base ($e\approx 2.718>0$) raised to any power, and a positive number to any real power stays positive. For negative $x$ we get $e^{-x}=\frac{1}{e^x}$ — a positive divided by a positive — still positive, just small. So the range is $y>0$; the graph hugs the line $y=0$ but never touches it.

L1.3 $\boxed{1}$. This is the defining property of the base $e$: among all curves $a^x$, the base $e$ is the one whose slope at the $y$-axis is exactly $1$.

L1.4 $(1,e)\approx (1,2.718)$. Why: put $x=1$ into $e^x$ to get $e^1=e$.

L2.1 $e^2=e\cdot e\approx (2.718{)}^2\approx 7.389$. Why: the index law $e^2=e^1\cdot e^1$. $e^{-2}=\frac{1}{e^2}\approx \frac{1}{7.389}\approx 0.135$. Why: a negative exponent means reciprocal.

L2.2 The slope of $e^x$ at any point is $e^x$ itself (slope = height). At $x=3$ the slope is $e^3=e\cdot e\cdot e\approx 20.09$.

L2.3 Treat this as an inside-then-outside composite. The inside is $5x$; the outside is "$e$ to the something". Differentiating $e^{\text{something}}$ gives $e^{\text{something}}$ back, then we multiply by the derivative of the inside ($5$): $\frac{dy}{dx}=5e^{5x}.$ (This is the Chain rule in action; the parent note previews it.)

L2.4 The constant $4$ just rides along: $\frac{dy}{dx}=4\cdot \frac{d}{dx}{e}^x=4e^x.$

L3.1 Point: $(0,e^0)=(0,1)$. Slope there: $e^0=1$. Point–slope form gives $y-1=1(x-0)⟹y=x+1.$ Estimate at $x=0.1$: $y=0.1+1=1.1$. True value: $e^{0.1}\approx 1.10517$. The tangent under-guesses by only about $0.005$ — tiny, because near $x=0$ the curve and its tangent line almost overlap. This is exactly why $e^x\approx 1+x$ for small $x$. See figure below.

L3.2 Slope = height, so we need the height to be $10$: $e^x=10.$ To undo "$e$ to the power", we use the natural logarithm, the inverse of $e^x$: $x=\ln 10\approx \mathrm{2.3026.}$ The point is $(\ln 10,10)\approx (2.303,10)$.

L3.3 Slope = height, and the curve is higher at $x=1$ (height $e\approx 2.718$) than at $x=-1$ (height $e^{-1}\approx 0.368$). So it is steeper at $x=1$. Confirming: slope$(1)=e\approx 2.718$ vs slope$(-1)=e^{-1}\approx 0.368$. Steeper at $x=1$, as predicted.

L4.1 Inside is $-x$ (slope $-1$); outside is $e^{(\cdot )}$. Chain rule: $\frac{dy}{dx}=e^{-x}\cdot (-1)=-e^{-x}.$ At $x=0$: slope $=-e^0=-1$. Meaning: the slope is negative, so $y=e^{-x}$ is a decreasing curve — it falls as $x$ increases. (It's the mirror image of $e^x$ across the $y$-axis.)

L4.2 Point: $(0,e^0)=(0,1)$. Slope: $\frac{d}{dx}{e}^{2x}=2e^{2x}$, and at $x=0$ that is $2e^0=2$. Point–slope form: $y-1=2(x-0)⟹y=2x+1.$ Notice the slope is $2$, not $1$ — the "$\times 2$" from the chain rule steepens the tangent.

L4.3 Differentiate term by term: $\frac{dy}{dx}=e^x+(-e^{-x})=e^x-e^{-x}.$ At $x=0$: $e^0-e^0=1-1=0$. Geometric meaning: a zero slope means the tangent line is horizontal — the curve momentarily stops rising or falling. Here it's the bottom of a valley (a minimum): the two exponentials pull in opposite directions and exactly cancel at $x=0$. See figure.

L5.1 (a) $P(0)=500e^{0.4\cdot 0}=500e^0=500\cdot 1=500$ bacteria. The coefficient $500$ is the starting amount.

(b) Inside the exponent is $0.4t$ (slope $0.4$); the constant $500$ rides along. Chain rule: $\frac{dP}{dt}=500\cdot e^{0.4t}\cdot 0.4=200e^{0.4t}.$

(c) Rewrite using $P(t)=500e^{0.4t}$: $\frac{dP}{dt}=200e^{0.4t}=0.4\underset{P(t)}{\underbrace{\left(500e^{0.4t}\right)}}=0.4P(t).$ So the growth rate is $0.4$ times the current population — this is exactly the exponential growth signature "rate ∝ amount", with constant of proportionality $k=0.4$ per hour.

(d) At $t=5$: $P(5)=500e^{0.4\cdot 5}=500e^2\approx 500\cdot 7.389=3694.5\text{bacteria}.$ $\frac{dP}{dt}{|}_{t=5}=200e^2\approx 200\cdot 7.389=1477.8\text{bacteria per hour}.$ Check part (c): $0.4\times 3694.5=1477.8$ ✓ — the rate really is $0.4$ times the population.

L5.2 Set $P(t)=2000$: $500e^{0.4t}=2000⟹e^{0.4t}=4.$ Undo the $e$-exponent with the natural log (its inverse): $0.4t=\ln 4⟹t=\frac{\ln 4}{0.4}\approx \frac{1.3863}{0.4}\approx 3.466\text{hours}.$

L1 facts ::: $e^0=1$, range $y>0$, slope-$1$ at origin, passes $(1,e)$ L2 rules ::: $\frac{d}{dx}{e}^{kx}=k{e}^{kx}$; negative exponent = reciprocal L3 inverse ::: undo $e^x=c$ with $x=\ln c$ L4 chain + geometry ::: multiply by slope of the inside; zero slope = horizontal tangent L5 model ::: $P=P_0{e}^{kt}\Rightarrow \frac{dP}{dt}=kP$ (rate proportional to amount)