State True or False:

(a.) The function f, defined on R by,
            f(x) = ln |(1 - e^-x) / (1 + e^-x)|,
is an even function.

(b.) The tangent to the curve, x^3 + 6 y^2 + 5 x = 0, at (— 1, 1) is perpendicular to the y-axis.

(c.) [So here, the integral or int sign is S… And upper and lower limit will be in () = S (x,y)] [refer pic]

         d/dx {S (root x, 1) tan^2 (t^2)} = (tan^2) x / (root x) -1,
      for x member of ]1, infinity[

(d.) The graph of the function, y = x + Ix| for all x member of Q, is strictly increasing.

(e.) Every integrable function is differentiable.




.... Show steps and explain as much as you can.

in Calculus Answers by Level 2 User (1.3k points)

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1 Answer

a) An even function is such that f(x)=f(-x).




Therefore if f(x)=ln|(1-e⁻ˣ)/(1+e⁻ˣ)|, f(x) is even: TRUE

b) Differentiate:


At (-1,1) [which satisfies the equation of the curve, so the point is on the curve], we get: 3+12dy/dx+5=0, dy/dx=-⅔. This is perpendicular to neither axis: so FALSE.

c) d/dx{∫[1,√x](tan²(t²)dt)}=tan²(x)/(√x-1), where the integral is between a low limit of 1 and a high limit of √x and x∈[1,∞). If the equation were true for all values of x in the given range, then it has to be true for x=1. But that would make the right-hand side expression infinite, so it’s FALSE.

Also, d(F(t²))/dt=2tF'(t), that is, ∫2tF'(t)dt=F(t²), and applying the interval we would have F(x)-F(1), which differentiates to F'(x), because F(1) is a constant. But the term 2t is missing from the integrand which is just tan²(t²) and not 2ttan²(t²).

d) For x<0, y=0 and for x≥0, y=2x, where x∈ℚ, so y is only strictly increasing when x≥0; therefore FALSE.

e) f(x)=|x| is integrable for all x but not differentiable at x=0 because its gradient is not definable at x=0. Compare with f(x)=x which is integrable and everywhere differentiable (gradient=1). Integration can be defined as the area under a curve (or straight line for linear functions) and the area under f(x)=|x| is defined because it’s a continuous function, but it’s not differentiable everywhere. So: FALSE.

by Top Rated User (825k points)

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