When we are sketching the graph of functions either by hand or through any graphing software we cannot show the entire graph.

**Sketching of
Curves**

When we
are sketching the graph of functions either by hand or through any graphing
software we cannot show the entire graph. Only a part of the graph can be
sketched. Hence a crucial question is which part of the curve we need to show
and how to decide that part. To decide on this we use the derivatives of
functions. We enlist few guidelines for determining a good viewing rectangle
for the graph of a function. They are :

(i) The
domain and the range of the function.

(ii) The
intercepts of the cure (if any).

(iii) Critical
points of the function.

(iv) Local
extrema of the function.

(v) Intervals
of concavity.

(vi) Asymptotes
of the curve (if exists)

(vii) Points
of inflexions (if any).

** **

**Example 7.69**

Sketch
the curve *y* =
*f* (*x* ) = *x*^{2}
−
*x* − 6 .

**Solution**

Factorising
the given function, we have

*y*= *f* (*x* ) = (*x* − 3)(*x* + 2) .

(1) The
domain of the given function *f (x)* is
the entire real line.

(2) Putting* y *= 0 we get* x *= −2, 3 . Therefore the *x
-*intercepts are (−2, 0) and (3, 0) putting* x *= 0 we get* y *= −6 .
Therefore the* y *-intercept is (0, −6)
.

(3)* f *′(*x*)* *=* *2*x
*−1* *and hence the critical point of the
curve occurs at *x* =
1/2 .

(4) *f *′′(*x*)* *=* *2*
*>* *0,* *∀*x *. Therefore at* x *=* *1/2* *the curve has a local minimum which is* f*(1/2)* *= − 25/4.

(5) The
range of the function is* y *≥ − 25/4

(6) Since
*f* ′′(*x*)
=
2 >
0, ∀*x* the function is concave upward in
the entire real line.

(7) Since
*f* (*x*) = 2 ≠ 0, ∀*x* the curve has no points of
inflection.

(8) The
curve has no asymptotes.

The
rough sketch of the curve is shown on the right side.

** **

**Example 7.70**

Sketch
the curve *y *=* f *(*x
*)* *=* x*^{3}* *−* *6*x *−* *9* *.

**Solution**

Factorising
the given function, we have

* y*
= *f (x)* = (x − 3)(x^{2} + 3x
+ 3) .

(1) The
domain and the range of the given function *f
(x)* are the entire real line.

(2) Putting* y *= 0 , we get the* x *= 3 . The other two roots are imaginary. Therefore, the* x *-intercept is (3, 0) . Putting* x *= 0, we get* y *= −9. Therefore, the y-intercept is (0, −9) .

(3) *f *′*
*(*x*) = 3(*x*^{2} − 2) and hence the critical points of the curve occur
at* x *= ± √2 .

(4)* f *′′*(x)*
= 6x . Therefore at* x *= √2 the curve
has a local minimum because* f *′′ (√2)=6√2
> 0 . The local minimum is* f *(√2)=−4√2
− 9 . Similarly* x *= − √2 the curve
has a local maximum because* f *′′( − 2)=−6√2
< 0 . The local maximum is* f *(−√2)=4
√2−9.

(5) Since
*f *′′(* x *)* *=* *6*x *>* *0,* *∀* x *>* *0* *the function is
concave upward in the positive real line. As *f *′′(* x *)* *=* *6*x *<* *0,* *∀* x *<* *0* *the function is
concave downward in the negative real line.

(6) Since
*f* ′′ ( *x*
)
=
0 at *x* = 0 and *f* ′′( *x*) changes its
sign when passing through *x* =
0 . Therefore the point of inflection is ( 0, *f* ( 0)) = ( 0, −9) .

(7) The
curve has no asymptotes.

The
rough sketch of the curve is shown on the right side.

** **

**Example 7.71**

Sketch
the curve *y* =
*x*^{2}−3*x* / (*x*−1).

**Solution**

Factorising
the given function we have,

* y *= *f* (*x*) = *x*(*x*−3) / (*x*−1).

(1) The
domain and the range of *f* ( *x*) are respectively *R *\ {1}* *and the entire
real line.

(2) Putting
*y* = 0 we get the *x* = 0, 3 . Therefore the *x
*-intercept is* *(3, 0)* *. Putting* x *=* *0*
*, we get* y *=* *0* *.

Therefore the curve passes through the origin.

(3) *f* ′(*x*)
=*
x*^{2}−2*x*+3 / (x
−1)^{2}* *and hence
the critical point of the curve occurs at*
x *= 1 as *f *′ (1) does not exist.
But* x*^{2} − 2*x* + 3 = 0 has no real solution. Hence
the only critical point occurs at* x *=
1 .

(4) *x* = 1 is not in the domain of the
function and* f *′(x) ≠ 0 ∀* x *∈ **R**\{1}, there is no local maximum or local minimum.

(5)* f *′′ (*x*) = − 4 / (*x* −1)^{3} ∀x ∈ \{1}. Therefore when* x *< 1, *f*′′*(x)* > 0 the curve
is concave upwards in (−∞,1) and when* x *>
1, *f*′′* (x)* < 0 the curve is concave downwards in (1, ∞) . Since *x* = 1 is not in the domain *f*′′*(x)*
≠ 0* x *∈ **R**\{1} there is no point of infection for *f (x)* .

(6) Since,
lim_{ x→1−}* x*^{2}−3x /
(x−1) = +∞ and lim_{x→1+}*
x*^{2}−3*x* / (*x*−1) = −∞ ,* x *=1 is a vertical asymptote.

The
rough sketch is shown on the right side.

** **

Sketch
the graph of the function *y* = 3*x* / *x*^{2}
−1

(1) The
domain of *f* ( *x*) is ℝ
\ {−1,1}.

(2)
Since *f *(* *−* x*,* *−* y *)* *=* f *(* x *,*
y *)* *, the curve is symmetric about the origin.

(3) Putting
*y* = 0 , we get *x* = 0 . Hence the *x*
-intercept is ( 0, 0) .

(4) Putting
*x* = 0 , we get *y *=* *0* *. Hence the* y *-intercept is* *(* *0, 0)* *.

(5) To
determine monotonicity, we find the first derivative as *f* ′( *x* )
=
−3(*x*^{2}+1)
/ ( *x*^{2}−1)^{2}.

Hence, *f* ′ ( *x*
)
does not exist at *x* = −1,1.
Therefore, critical numbers are *x* = −1,1
.

The
intervals of monotonicity is tabulated in Table 7.9.

(6) Since
there is no sign change in *f* ′
(
*x* ) when passing through critical
numbers. There is no local extrema.

(7) To
determine the concavity, we find the second derivative as *f* ′′( *x*
)
= 6*x*(x^{2}+3) / (x^{2}−1)^{3}.

*f *′′(* x *)* *=* *0* *⇒* x *=* *0 and* f *′′* *(* x *)* *does not exist at* x *= −1,1* *.

The
intervals of concavity is tabulated in Table 7.10.

(8) As *x* = −1 and 1are not in the domain of *f* ( *x*
)
and at *x* =
0 ,the second derivative is zero and *f *′′(* x *)* *changes its sign from positive to negative when passing
through* x *=* *0* *. .Therefore, the
point of inflection is ( 0, *f* ( 0)) = ( 0, 0) .

(9)

Therefore
*x* = −1 and *x *=* *1* *are vertical
asymptotes.

The
rough sketch of the curve is shown on the right side.

Tags : Applications of Differential Calculus | Mathematics , 12th Maths : UNIT 7 : Applications of Differential Calculus

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12th Maths : UNIT 7 : Applications of Differential Calculus : Sketching of Curves | Applications of Differential Calculus | Mathematics

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