Step 2:
Find f '(x) and examine if it is defined at every point on the open interval (a, b). If f '(x) is defined for all x (a, b), then the function is differentiabl..
Find f '(x) and examine if it is defined at every point on the open interval (a, b). If f '(x) is defined for all x (a, b), then the function is differentiabl..Step 4:
If Mean value theorem is applicable, solve the equation Show that one of the roots lie in the open interval (a, b). This verifies the Mean Value Theore..
If Mean value theorem is applicable, solve the equation Show that one of the roots lie in the open interval (a, b). This verifies the Mean Value Theore..Step1:
We know that every polynomial function is continuous and product of continues functions are continuous. f (x), being product of polynomials of degree 1, is a continuous function in [4,10..
Step 3:
Since both the condition are satisfied, Mean Value Theorem is applicable. \ There exist c (4, 10) such that \ f (4) = 0 f '(c) = 3c 2 - 36c + 104 Substituting these values in (1), we have Since 8 (4,10), Mean Value Theorem is satisfi..
Since both the condition are satisfied, Mean Value Theorem is applicable. \ There exist c (4, 10) such that \ f (4) = 0 f '(c) = 3c 2 - 36c + 104 Substituting these values in (1), we have Since 8 (4,10), Mean Value Theorem is satisfi..Monotonic Function
A function is said to be monotonic if it is either increasing or decreasing but not both in a given interval. Consider the function The given function is increasing function on R. Therefore it is a monotonic function in [0,1]. It has its minimum value at x = 0 which is equal to f (0) =1, has a m..
A function is said to be monotonic if it is either increasing or decreasing but not both in a given interval. Consider the function The given function is increasing function on R. Therefore it is a monotonic function in [0,1]. It has its minimum value at x = 0 which is equal to f (0) =1, has a m..Working Rule for Finding Extremum Values Using First Derivative Test
Let f (x) be the real valued differentiable functio..
Step 2:
Solve f '(x) = 0 to get the critical values for f (x). Let these values be a, b, c. These are the points of maxima or minima. Arrange these values in ascending ord..
Step 4:
Let us take the critical value x= a. Find the sign of f '(x) for values of x slightly less than a and for values slightly greater than a. (i) If the sign of f '(x) changes from positive to negative as x increases through a, then f (a) is a local maximum value. (ii) If the sign of f '(x) changes fr..
Solution:
f (x) = x 3 - 6x 2 + 9x + 15 f ' (x) = 3x 2 -12x + 9 = 3(x 2 - 4x + 3) = 3 (x - 1) (x - 3) Thus x = 1 and x = 3 are the only points which could be the points of local maxima or local minima. Let us examine for x=1 When x<1 (slightly less than 1) f '(x) = 3 (x - 1) (x - 3) = (+ ve) (- ve)..
f (x) = x 3 - 6x 2 + 9x + 15 f ' (x) = 3x 2 -12x + 9 = 3(x 2 - 4x + 3) = 3 (x - 1) (x - 3) Thus x = 1 and x = 3 are the only points which could be the points of local maxima or local minima. Let us examine for x=1 When x<1 (slightly less than 1) f '(x) = 3 (x - 1) (x - 3) = (+ ve) (- ve)..See what our Users say :
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