Exercise 1 · Piecewise-defined function — continuity, area and tangent line

Continuity of a piecewise-defined function, computing an area with integrals and a tangent line with a given slope.

Maximum score · 2.5 points

Consider the piecewise-defined function $$f(x)=\begin{cases} 5e^{2x} & x\le 0 \\ (x+m)^2+1 & 0

Find the values of $m$ that make the function $f(x)$ continuous on its whole domain. Justify your answer. 1 p
Sketch the graph of $y=f(x)$ for the case $m=-2$, and compute the area bounded by this graph, the $OX$ axis and the lines $x=-1$ and $x=3$. 1 p
For $m=-2$, find a point where the tangent line to $y=f(x)$ is parallel to $y=-2x$. Compute the equation of this tangent line. 0.5 p

Science Bachillerato · Block D — Analysis Continuity Definite integral Tangent line

Step-by-step solution

Key idea: a piecewise function is continuous at a junction point when both one-sided limits agree with the value of the function. Each junction gives one condition on $m$; all of them must hold at the same time.

a) Values of $m$ that make $f$ continuous

Within each piece the function is continuous (exponential, polynomial and constant). We only need to study the two junction points, $x=0$ and $x=2$.

At $x=0$: the left-hand limit is $5e^{0}=5$ and the right-hand limit is $(0+m)^2+1=m^2+1$. We impose equality:

m^2+1=5 \;\Longrightarrow\; m^2=4 \;\Longrightarrow\; m=-2 \;\text{ or }\; m=2.

At $x=2$: the left-hand limit is $(2+m)^2+1$ and the value (and right-hand limit) is $1$. We impose equality:

(2+m)^2+1=1 \;\Longrightarrow\; (2+m)^2=0 \;\Longrightarrow\; m=-2.

Both conditions must hold at the same time. The value $m=2$ passes the first one but fails at $x=2$; the only common value is $m=-2$.

$f$ is continuous on its whole domain only if $m=-2$.

b) Sketch and area for $m=-2$

With $m=-2$ the middle piece is $(x-2)^2+1$. The graph: an increasing exponential up to $(0,5)$, a parabola descending from $(0,5)$ to the vertex $(2,1)$ and, from there on, the constant line $y=1$. Since $f(x)>0$ everywhere, the area is directly the integral of $f$ between $x=-1$ and $x=3$:

We split the integral into the three pieces:

\text{Area}=\int_{-1}^{0} 5e^{2x}\,dx+\int_{0}^{2}\big((x-2)^2+1\big)\,dx+\int_{2}^{3} 1\,dx.

First piece:

\int_{-1}^{0} 5e^{2x}\,dx=\Big[\tfrac{5}{2}e^{2x}\Big]_{-1}^{0}=\tfrac{5}{2}\big(1-e^{-2}\big)=\tfrac{5}{2}-\tfrac{5}{2}e^{-2}.

Second piece (with the substitution $u=x-2$):

\int_{0}^{2}\big((x-2)^2+1\big)\,dx=\Big[\tfrac{(x-2)^3}{3}+x\Big]_{0}^{2}=\Big(0+2\Big)-\Big(-\tfrac{8}{3}+0\Big)=2+\tfrac{8}{3}=\tfrac{14}{3}.

Third piece: $\int_{2}^{3}1\,dx=1$. Adding up:

\text{Area}=\Big(\tfrac{5}{2}-\tfrac{5}{2}e^{-2}\Big)+\tfrac{14}{3}+1=\frac{49}{6}-\frac{5}{2}e^{-2}\approx 7{,}83\ \text{u}^2.

Area $=\dfrac{49}{6}-\dfrac{5}{2}e^{-2}\approx 7{,}83$ square units.

c) Tangent line parallel to $y=-2x$

"Parallel to $y=-2x$" means slope $-2$, that is, we look for a point where $f'(x)=-2$. We differentiate piece by piece (with $m=-2$):

f'(x)=\begin{cases} 10e^{2x} & x<0 \\ 2(x-2) & 02 \end{cases}

The exponential piece gives $10e^{2x}>0$ always (never $-2$) and the constant piece gives $0$. Only the middle piece remains:

2(x-2)=-2 \;\Longrightarrow\; x-2=-1 \;\Longrightarrow\; x=1\in(0,2).\ \checkmark

The point of tangency is $\big(1,f(1)\big)$ with $f(1)=(1-2)^2+1=2$. The line of slope $-2$ through it:

y-2=-2(x-1)\;\Longrightarrow\; y=-2x+4.

Point $(1,2)$; tangent line $y=-2x+4$.