#### To determine

**To find**:

a) g'0 and g''(0) in terms of c.

b) Equation of tangent line to the graph of f at x=0.

#### Answer

(a) g''0=c2-2

(b) y=5+3kx+3

#### Explanation

a) g'x=cecx+f'(x) implies g'0=c+5. Also g''x=c2ecx+f''(x), then g''0=c2-2.

b) h'x=ekxf'x+k ekxfx implies h'0=f'0+kf0=5+3k. You know that the slope of tangent line is 5+3k. Therefore, (0,3) is the point on the tangent line, when x=0. Now you have a point on the tangent line and the slope of the tangent line. The only step left is to use the point (0,3) and the slope 5+3k in the point slope formula for a line. Therefore,

y-y1=m(x-x1), i.e., y-3=5+3k(x-0), i.e.,y=5+3kx+3. This is the equation for the tangent line in terms of k.

**Conclusion:**

(a) g''0=c2-2

(b) y=5+3kx+3