Answer:
Explanation:
Given that v(x) = g(x)×(3/2*x^4+4x-1)
Let's find V'(2)
V(x) is a product of two functions
● V'(x) = g'(x)×(3/2*x^4+4x-1)+ g(x) ×(3/2*x^4+4x-1)
We are interested in V'(2) so we will replace x by 2 in the expression above.
g'(2) can be deduced from the graph.
● g'(2) is equal to the slope of the tangent line in 2.
● let m be that slope .
● g'(2) = m =>g'(2) = rise /run
● g'(2) = 2/1 =2
We've run 1 square to the right and rised 2 squares up to reach g(2)
g(2) is -1 as shown in the graph.
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Let's derivate the second function.
Let h(x) be that function
● h(x) = 3/2*x^4 +4x-1
● h'(x) = 3/2*4*x^3 + 4
● h'(x) = 6x^3 +4
Let's calculate h'(2)
● h'(2) = 6 × 2^3 + 4
● h'(2) = 52
Let's calculate h(2)
●h(2) = 3/2*2^4 + 4×2 -1
●h(2)= 31
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Replace now everything with its value to find V'(2)
● V'(2) = g'(2)×h(2) + g(2)× h'(2)
● V'(2)= 2×31 + (-1)×52
●V'(2) = 61 -52
●V'(2)= 9