logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

Find a point on the curve $y=(x-3)^2$,where the tangent is parallel to the chord joining the point (3,0) and (4,1).

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Let $f(x)$ be a function defined on $[a,b]$ such that
  • (i) it is continuous on $[a,b]$
  • (ii) it is differentiable on $(a,b)$
  • Then,there exists $c\in (a,b)$ such that $f'(c)=\large\frac{f(b)-f(a)}{b-a}$
Step 1:
$y=(x-3)^2$------(1)
The given points are $(3,0)$ and $(4,1)$.
Let $y=f(x)$
Then $f(x)=(x-3)^2,x\in[3,4]$
Since $f(x)$ is a polynomial function $f(x)$ is continuous in $[3,4]$
$f'(x)=2(x-3)$ which exists in $(3,4)$
Therefore it is also differentiable in $(3,4)$
Hence both the conditions of Lagrange's Mean value theorem is satisfied.
Step 2:
$f(3)=0$
$f(4)=1$
$f'(c)=2(c-3)$
Therefore $2c-6=\large\frac{1-0}{4-3}$
$\Rightarrow 2c-6=1$
Therefore $c=\large\frac{7}{2}$
$\large\frac{7}{2}$$\in (3,4)$
Since $c$ is the $x$-coordinate of that point at which the tangent is parallel to the chord joining the points $(3,0)$ and $(4,1)$.
Therefore putting $x=\large\frac{7}{2}$ in equ(1) we get,
$y=\big(\large\frac{7}{2}-$$3)^2$
$\;\;=\large\frac{1}{4}$
Hence the required points are $\big(\large\frac{7}{2},\frac{1}{4})$
answered Jul 3, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...