Math Assignment Class VIII | Square & Square Root

Basic concepts, definitions and formulas of mathematics, mathematics assignments for 9th standard to 10+2 standard, maths study material for 8th, 9th, 10th, 11th, 12th classes, Mathematics lesson plan for 10th and 12th standard, Interesting maths riddles and maths magic, Class-wise mathematics study material for students from 9th to 12
Concept of increasing and decreasing and their graphical
representation. |
(i) Increasing on I if x1 <
x2 on I
(ii) Strictly increasing on I if x1 < x2 on I
(iii) Decreasing on I if x1 < x2 on I
(iv) Strictly Decreasing on I if x1 < x2 on I
Conclusions
Critical Points :
RESULTS
1) f (x) is strictly increasing in (a, b), if f '(x) > 0 for each x ϵ (a, b) Use of Baby Curve Method
in increasing and decreasing of functions Steps to be followed:
Note: We cannot change the sign at the critical point whose
factor is with the even power. Example: If any factor of f '(x) is (x - 2)2 or
(x - 3)4 , then we cannot change the sign on the baby curve
at point 2 and 3. Here we maintain the same sign . Power of (x - 3) is an even number so here
sign remain unchanged. |
Equation of Tangent and Equation of Normal to the Curve |
Tangent to a curve:
It is a line which just touches the curve at one point.
Normal line:
At every tangent point there is a line which is perpendicular to the tangent line is called a normal line.
Equation of Tangent
Equation of the tangent at (x1 , y1) is given by
y – y1 = m(x - x1) where m is the slope of the tangent
To find the approximate value we use f(x + Δx) = f(x) + f '(x)Δx
Explanation through an example\[\sqrt{25.3}=\sqrt{25+0.3}=\sqrt{x+\Delta x}\]\[\left (25.3 \right )^{1/2}=\left ( 25+0.3 \right )^{1/2}=\left ( x+\Delta x \right )^{1/2}\]\[let\; y\; =x^{1/2}=\left ( 25 \right )^{1/2}=5\; and\; \Delta x=0.3\]\[\frac{dy}{dx}=\frac{1}{2x^{1/2}}=\frac{1}{2\left ( 25 \right )^{1/2}}=\frac{1}{10}\]\[f(x+\Delta x)=f(x)+f'(x)\Delta x\]\[(x+\Delta x)^{1/2}= y+\frac{dy}{dx}\times \Delta x\]\[(25+0.3)^{1/2}= 5+\frac{1}{10}\times 0.3\]\[(25.3)^{1/2}= 5+0.03 = 5.03\]
Concept of Maxima and Minima
Maximum Value:
Let f(x) be a real function defined on an interval I. Then f(x) is said to have the maximum value in I, if there exist a point c in I such that f(x) ≤ f(c) for all x ϵ I
In this case f(c) is called the maximum value
Minimum Value :
Let f(x) be a real function defined on an interval I. Then f(x) is said to have the maximum value in I, if there exist a point c in I such that f(x) ≥ f(c) for all x ϵ I
In this case f(c) is called the minimum value
Modulus function f(x) have minimum value at x = 0. But f(x) do not have maximum value over the set of real numbers. But if we restrict the interval [-3, 1] then it has maximum value at x = -3
Note 1 : Every monotonic function attains maximum/minimum value only at the end points of the domains(or in close interval). If interval is open then the monotonic function do not have maximum / minimum.
Monotonic Function:
Monotonic Function means the function which is either increasing or decreasing in their entire domain or interval e.g. : f(x) = x is a monotonic function. In the above figure f(x) = x do not have maximum or minimum value in (0, 4). If there is a closed interval [0, 4], then f(x) is minimum at x = 0 and f(x) is maximum at x = 4.On the x-axis as we move from left to right g(x) is continuously goes on increasing. It means that g(x) is a monotonic increasing function on real numbers.
Note 2 :
Every continuous function on a closed interval has maximum and minimum value.
Local Maximum:
A function f(x) is said to attain a local maximum at x = a if there exist a neighborhood (a - 𝛿, a + 𝛿) of a such that
f(x) < f(a) for all x ϵ (a - 𝛿, a + 𝛿) here f(a) is called the local maximum value of f(x)
First Derivative Test for finding Maxima and Minima
Let f be a function defined on an open interval I. Let f is continuous and differentiable at the critical point in interval I then follow the following Algorithm.
iv) The critical point at which f '(x) changes its sign from negative (- ve) to positive (+ ve) (as we move from left to right) is called the point of local minima and the function have local minimum value at this point.
v) The critical point at which f '(x) does not change its sign is called the point of inflection.
vi) If f '(x) = 0, then First Derivative Test Fails. In this case we can discuss the maxima and minima by using definition.
Second Derivative Test of finding Maxima & Minima
Let f be a function defined on an open interval I. Let f is continuous and twice differentiable at the critical point in interval I then follow the following Algorithm.
i) Find f '(x) and put f '(x) = 0 to find the critical points.
ii) Find f ''(x) and find the value of f''(x) at the critical points.
iii) If f ''(c) < 0 , then f(x) has local maximum value at x = c
iv) If f ''(c) > 0, then f(x) has local minimum value at x = c
v) If f '' (c) = 0 then test fail.
Note: If second derivative test fails then we can find the maximum and minimum value by using first derivative test. (OR)
Find f ''' (x), If f ''' (c) ≠ 0, then x = c is the point of inflection
Minimum and maximum value of a function in closed interval
We know that the function like f(x) = x + 5 is a monotonic function and is neither have maximum value, nor have minimum value in any open interval like (0, 3). This function also neither have local maximum value nor have local minimum value. But if we extend the open interval to close interval [0, 3], then this function have maximum value at x = 3 and minimum value at x = 0 . This maximum value is called Absolute Maximum value and minimum value is called absolute minimum value.
Let us suppose that a function f(x) defined in the closed interval [a, d]
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