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 \(x = 2\).
Answer
Finally, we substitute \(x = 2\) into \(h'(x)\) to find the slope of the tangent line at \(x = 2\). So, the slope is \(h'(2) = 3 \cdot 2(2 - 2) + 2 = 2\).
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, \(h'(x)\), is \(3 \cdot 2(x - 2) + 2\).
Step 3 ：Finally, we substitute \(x = 2\) into \(h'(x)\) to find the slope of the tangent line at \(x = 2\). So, the slope is \(h'(2) = 3 \cdot 2(2 - 2) + 2 = 2\).
