gx3x2 hxx3 hgx and hgx and evaluate hg1 Suppose that the fun

g(x)=3x²

h(x)=x³

(h*g)(x) and (h-g)(x) and evaluate (h+g)(1)

Solution

Given: g(x) = 3x²

h(x) = x³

(h*g)(x) = h(x)*g(x)

putting the values of h(x) and g(x) in the above, we get:

(h*g)(x) = (x³)*(3x²)

= 3x⁵

Hence, (h*g)x =  3x⁵

(h-g)*(x) = h(x) - g(x)

putting the values of h(x) and g(x) in the above, we get:

(h-g)(x) = x³ - 3x²

Now simplify (h+g)(-1)

we know that (h-g)(x) = x³ - 3x²

simply we can calculate for (h+g)(x),

(h+g)(x) = h(x) + g(x)

putting the values of h(x) and g(x) in the above, we get:

(h+g)(x) = x³ + 3x².........(1)

now simplify (h+g)(-1)

from the equation (1), putting x = -1, we get:

(h+g)(-1) = (-1)³ + 3*(-1)²

(h+g)(-1) = -1+ 3*(1)

(h+g)(-1) = -1+ 3

Hence, (h+g)(-1) = 2