Chapter 4, Problem 81RE

a.

To determine

To define: a function g by $g(x)=e^{kx}+f(x)$ ,where k is a constant. Find g’(0) and g’’(0) in term of k.

Answer to Problem 81RE

  $\begin{array}{l}g\text{'}(0)=k+3.\\ g\text{'}\text{'}(0)=k^2-1.\end{array}$

Explanation of Solution

Given information: A function f and its first and second derivatives are defined for all

real numbers ,and it is given that f(0)=2,f’(0)=3 and f’’(0)=-1.

Calculation:

  $\begin{array}{l}g(x)=e^{kx}+f(x).\\ g\text{'}(x)=e^{kx}.k+f\text{'}(x).\\ g\text{'}(x)=k{e}^{kx}+f\text{'}(x).\\ g\text{'}(0)=k{e}^0+f\text{'}(0)=k(1)+3=k+3.\\ g\text{'}\text{'}(x)=k{e}^{kx}.k+f\text{'}\text{'}(x).\\ g\text{'}\text{'}(x)=k^2{e}^{kx}+f\text{'}\text{'}(x).\\ g\text{'}\text{'}(0)=k^2{e}^0+f\text{'}\text{'}(0)=k^2(1)+(-1)=k^2-1.\end{array}$

b.

To determine

To define: a function h by $h(x)=\cos (bx)f(x),$ where b is a constant. Find h’(x)and write an equation for the line tangent to the graph of h at x =0.

Answer to Problem 81RE

Tangent line equation is $y=3x+2.$

Explanation of Solution

Given information: A function f and its first and second derivatives are defined for all

real numbers ,and it is given that f(0)=2,f’(0)=3 and f’’(0)=-1.

Formula used:

The equation of a line in point slope form is

  $y-y_1=m(x-x_1).$ Where m is the slope and $(x_1,y_1)$ is the point of the line.

Calculation:

  $\begin{array}{l}h(x)=\cos (bx)f(x).\\ h\text{'}(x)=-\sin (bx).b.f(x)+\cos (bx).f\text{'}(x).\\ h\text{'}(x)=-b\sin (bx).f(x)+\cos (bx).f\text{'}(x).\\ h\text{'}(0)=-b\sin (0)f(0)+\cos (0).f\text{'}(0)=0+1(3)=3.\\ h(x)=\cos (bx)f(x).\\ h(0)=\cos (0)f(0)=1(2)=2.\end{array}$

Substitute the point (2,0) and slope m =3 into the point-slope form of a line to write the equation of the tangent line.

  $\begin{array}{l}y-2=3(x-0).\\ y-2=3x.\\ y=3x+2.\end{array}$