Splet16. mar. 2024 · Find also the local maximum and the local minimum values, as the case may be: (iv) f (𝑥)=sin𝑥 –cos𝑥, 0 0 f’’ (𝑥)>0 when 𝑥 = 7𝜋/4 Thus 𝑥 = 7𝜋/4 is point of local minima f (𝑥) has minimum value at 𝑥 = 7𝜋/4 As sin (2π − 𝜃) = –sin 𝜃 & cos (2π − 𝜃) = cos 𝜃 Local minimum value is f (𝑥)=𝑠𝑖𝑛𝑥−𝑐𝑜𝑠𝑥 f (7𝜋/4)=𝑠𝑖𝑛 (7𝜋/4)−𝑐𝑜𝑠 (7𝜋/4) =𝑠𝑖𝑛 (2𝜋−𝜋/4)−𝑐𝑜𝑠 (2𝜋−𝜋/4) … SpletAOD (TN) - Free download as PDF File (.pdf), Text File (.txt) or read online for free. APPLICATION OF DERIVATIVE Syllabus in IIT : Tangents and normals, increasing and decreasing functions, maximum and minimum values of a function, applications of Rolle's theorem and Lagrange's Mean value theorem. Tangent-Normal-Rate Measure & …
MODEL QUESTION PAPER ¼ mPp xf
SpletSince the 3 functions have coefficient of x as 1, therefore their periods are same (2pi). 0.785 = pi/4, therefore, maxima will occur at pi/2-pi/4=pi/4. So it is indicated in terms of pi in the graph. Whereas, 1.107 or 0.464 are not a common fraction of pi. So, its maxima will be present at some decimal value, which is not a common fraction in ... Splet02. apr. 2016 · calculus - Maximum value of $f (x) = \cos x \left ( \sin x + \sqrt {\sin^2x +\sin^2a}\right)$ - Mathematics Stack Exchange Maximum value of f(x) = cosx(sinx + √sin 2x + sin 2a) Ask Question Asked 7 years ago Modified 7 years ago Viewed 4k times 8 Can we find maximum value of f(x) = cosx(sinx + √sin2x + sin2a) where ' a ' is a given constant. buddug humphreys
calculus - Maximum value of $f (x) = \cos x \left ( \sin x + \sqrt ...
Splet15. jun. 2024 · The equality occurs for x = y = 45 ∘, which says that we got a maximal value. In another hand, sin x + cos x = ( sin x + cos x ) 2 = 1 + sin 2 x ≥ 1. The equality occurs for x = 90 ∘, which says that we got a minimal value and since Y is a continuous function, we got the answer: [ 1, 2]. Share Cite answered Jun 15, 2024 at 20:10 SpletFind the absolute extrema of f(x)=sinx+cosx on [0,pi] Splet01. sep. 2016 · The maximum is obtained when tan x = 2, with x in Q1. And this is 2sin x + cos x , with tan x = 2 = 2( 2 √5) + 1 √5 = 5 √5 = √5. Of course, the minimum is −√5. Alternative method sans differentiation: f = √5(( 2 √5)sinx + ( 1 √5)cosx) = √5sin(x + α), where sinα = 2 √5 and cosα = 1 √5 M axf = √5 max sin(x +α) = √5(1) ' Answer link buddu broadcasting services