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A continuous function may be defined as a function that does not have any abrupt changes i.e. there are no discontinuities in the function.A function is said to be continuous if a small change in its output can be assured by restricting to a small change in its input.
In differential calculus, the derivative is a method of representing the instantaneous rate of change. that is, the amount by which a function is changing at one given point. For functions that act on the real numbers, it is the slope of the tangent line at a point on a graph.
Literal meaning of Integral is to include something. In mathematics, integral may be defined as assigning the numbers to functions in a way that describes displacement, area, volume, and other concepts that arise by combining infinitesimal data. The method of finding integrals is called integration.
In mathematics, limit may be defined as the value that a function approaches as the input approaches some value. This is represented by limits in a function and are essential to calculus and mathematical analysis, and are used to define continuity, derivatives, and integrals.
In mathematics, a partial derivative of a function having tow or more than two variables may be defined as its derivative with respect to one of those variables, with the others held constant at the same time. Partial derivatives finds its application in vector calculus and differential geometry.
Q. What are the maximum and minimum of the function f(x)=x^3-5x^2-8x+24?
Sol. In order to find the extreme value first we have to find at which point does it have extreme value
f(x)=x^3-5x^2-8x+24
f^' (x)=3x^2-10x-8
So, at the extreme points value of first derivative is zero
f^' (x)=0
3x^2-10x-8=0
3x^2-12x+2x-8=0
3x(x-4)+2(x-4)=0
(x-4)(3x+2)=0
⇒either x-4=0 or 3x+2=0
⟹ x=4 or x= -2/3
Now, we have find at which we have maximum and at which we have minimum.
For that lets take the second derivative of function
f^'' (x)=6x-10
Now, at x=4
f^'' (4)=6(4)-10
=24-10
=14>0
Which means at x=4 we have minimum value of function
So, minimum value of the function will be
f(4)=(4)^3-5(4)^2-8(4)+24
=64-80-32+24
=-24
Now, at x= -2/3
f^'' (-2/3)=6 (-2/3) -10
=-4-10
=-14<0
Which means at x= -2/3 we have maximum value of the function.
So, maximum value of the function will be
f(-2/3)=(-2/3)^3-5(-2/3)^2-8(-2/3)+24
=-8/27-20/9+16/3+24
=724/27 = 26 22/27
Q. Find the equation of Normal of curve f(x)=x^3-4x+3 at where curve meet y axis.
Sol. First of all we have to find the point at which we want to find the equation of tangent.
f(x)=x^3-4x+3
f(x)=0-0+3
We have to find equation of normal at (0,3)
So, now In order to find slope of normal first we have to find the slope of tangent
f^' (x)_((0,3))=3x^2-4
f^' (x)_((0,3) )=0-4=-4
So, slope of normal=1/(slope of tangent)
slope of normal=-1/4
Now Equation of is given by, y=mx+c , Where m is the slope of the normal
So, equation will be y=-1/4 x+c
In order to find value of 'c' we have to satisfy the equation of the line with the given point.
3=-1/4+c
c=7/4
y=-1/4 x+7/4
x+4y=7
So this is the equation of normal line at (0,3)
Q. Using limit definition of derivative find derivative of f(x)=sinx
Sol. We have f(x)=sinx
So, f(x+h)=sin(x+h)
f(x+h)-f(x)=sin(x+h)-sinx
Using sinx-siny=2 cos((x+y)/2) sin((x-y)/2)
f(x+h)-f(x)=2 cos((2x+h)/2) sin(h/2)
So now
f^' (x)=lim┬(h→0)〖(f(x+h)-f(x))/((x+h)-x)〗
f^' (x)=lim┬(h→0)〖(2 cos((2x+h)/2) sin(h/2) )/h〗
Using limit formula sinh/h=1
f^' (x)=lim┬(h→0)cos((2x+h)/2)
f^' (x)=lim┬(h→0)cos((2x+0)/2)
f^' (x)=cosx
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