Given the functions (f(x) = 3x^2 + 2x - 1) and (g(x) = x - 2), find the slope of the line tangent to the curve of the function (h(x) = f(g(x))) at (x = 2).

Question

Accepted Answer

Step:1 ：First, we need to find the function (h(x) = f(g(x))). Substituting (g(x)) into (f(x)), we have (h(x) = 3(g(x))^2 + 2g(x) - 1 = 3(x - 2)^2 + 2(x - 2) - 1).Step:2 ：Next, we find the derivative of (h(x)) using the chain rule. The derivative, (h&#x27;(x)), is (3 cdot 2(x - 2) + 2).Step:3 ：Finally, we substitute (x = 2) into (h&#x27;(x)) to find the slope of the tangent line at (x = 2). So, the slope is (h&#x27;(2) = 3 cdot 2(2 - 2) + 2 = 2).

Answer

Step: 1 ：First, we need to find the function (h(x) = f(g(x))). Substituting (g(x)) into (f(x)), we have (h(x) = 3(g(x))^2 + 2g(x) - 1 = 3(x - 2)^2 + 2(x - 2) - 1).Step: 2 ：Next, we find the derivative of (h(x)) using the chain rule. The derivative, (h&#x27;(x)), is (3 cdot 2(x - 2) + 2).Step: 3 ：Finally, we substitute (x = 2) into (h&#x27;(x)) to find the slope of the tangent line at (x = 2). So, the slope is (h&#x27;(2) = 3 cdot 2(2 - 2) + 2 = 2).

Given the functions $f (x) = 3 x^{2} + 2 x - 1$ and $g (x) = x - 2$ , find the slope of the line tangent to the curve of the function $h (x) = f (g (x))$ at $x = 2$ .

Answer

Popular Problems

Given the functions f(x)=3x2+2x−1 and g(x)=x−2, find the slope of the line tangent to the curve of the function h(x)=f(g(x)) at x=2.

Answer

Popular Problems

Given the functions $f (x) = 3 x^{2} + 2 x - 1$ and $g (x) = x - 2$ , find the slope of the line tangent to the curve of the function $h (x) = f (g (x))$ at $x = 2$ .