Hyperbolic function. & mean value theorem.

[ffuh205]

For For the hyperbolic function sinh(x^3+7) Find the value of c which verifies the mean value theorem where a = -1 and b = 1. This is what I have:

f'(c)=(f(b)-f(a))/(b-a); f'(c)=(f(1)-f(-1))/(1--1); f'(c)=(f(1)-f(-1))/2;
f(1) = [(e^(1+7) - (e^(-1+7)]/2 = 1288.76
f(-1) = [(e^(-1+7) - (e^(1+7)]/2 = -1288.76
f'(c)=( 1288.76 - - 1288.76)/2; f'(c)= 1288.76; 3x^2cosh(x^3+7) = 1288.76

I don't know how to solve for f'(c) at 1288.76 assuming I am correct up to that point

 

[ffuh205]

For this same problem, I also need to find the tangent line approximation at x = 1 and plot it on the graph of the function. This is what I have.
[(e^(1+7) - (e^(-1+7)]/2 = 1288.76; at (1, 1288.76) m = 3^2cosh(1^3+7); m = [e(7) + e(6)]/2 = 750.03; y - 1288.76 = 750.03(x – 1); y = 750.03x + 538.73

 

[BigGlenntheHeavy]

\(\displaystyle Given: \ f(x) \ = \ sinh(x^{3}+7), \ -1 \ \le \ x \ \le \ 1\)

\(\displaystyle f'(c) \ = \ \frac{f(1)-f(-1)}{2} \ = \ \frac{sinh(8)-sinh(6)}{2} \ = \ 644 \ = \ m\)

\(\displaystyle D_x[sinh(x^{3}+7)] \ = \ 3x^{2}cosh(x^{3}+7) \ = \ m\)

\(\displaystyle 3x^{2}cosh(x^{3}+7) \ = \ 644, \ x \ = \ .57 \ (Trusty \ TI-89)\)

\(\displaystyle f(.57) \ = \ 660, \ ergo, \ y-660 \ = \ 644(x-.57), \ \implies \ y \ = \ 644x+293\)

\(\displaystyle Hence, \ f'(.57) \ = \ 644, \ also \ there \ is \ another \ two, \ but \ I've \ omitted \ them.\)

\(\displaystyle See \ graph\)

[attachment=0:2uvazz6f]stu.jpg[/attachment:2uvazz6f]

 

[ffuh205]

How would I go about finding the inverse function of sinh (x^3 + 7)? I know that you need to transpose the y & x to solve for y but I am not sure how to get past a certain point. y = [e^(x^3 + 7) - e^-(x^3 + 7)]/2; 2y = e^(x^3 + 7) - e^-(x^3 + 7);
ln2y = (x^3 + 7) - - (x^3 + 7); ln2y = 2x^3 + 14; ln2y – 14 = 2x^3; (ln2y + 14)/2 = x^3
3?[(ln2y + 14)/2] = x; then y = 3?[(ln2x + 14)/2] or y = [(ln2x + 14)/2]1/3
Is this correct?

 

[galactus]

How would I go about finding the inverse function of sinh (x^3 + 7)? I know that you need to transpose the y & x to solve for y but I am not sure how to get past a certain point. y = [e^(x^3 + 7) - e^-(x^3 + 7)]/2; 2y = e^(x^3 + 7) - e^-(x^3 + 7);
ln2y = (x^3 + 7) - - (x^3 + 7); ln2y = 2x^3 + 14; ln2y – 14 = 2x^3; (ln2y + 14)/2 = x^3
3?[(ln2y + 14)/2] = x; then y = 3?[(ln2x + 14)/2] or y = [(ln2x + 14)/2]1/3
Is this correct?

By inverse, I assume you mean solve for x.

\(\displaystyle y=sinh(x^{3}+7)\)

\(\displaystyle sinh^{-1}(y)=x^{3}+7\)

\(\displaystyle x=(sinh^{-1}(y)-7)^{\frac{1}{3}}\)

\(\displaystyle y=(sinh^{-1}(x)-7)^{\frac{1}{3}}\)