the function $g$ is differentiable and satisfies $g(-1)=4$ and $g(-1)=2$. what is the approximation of…

the function $g$ is differentiable and satisfies $g(-1)=4$ and $g(-1)=2$. what is the approximation of $g(-1.2)$ using the line tangent to the graph of $g$ at $x = - 1$?\n(a) 3.6\n(b) 3.8\n(c) 4.2\n(d) 4.4

the function $g$ is differentiable and satisfies $g(-1)=4$ and $g(-1)=2$. what is the approximation of $g(-1.2)$ using the line tangent to the graph of $g$ at $x = - 1$?\n(a) 3.6\n(b) 3.8\n(c) 4.2\n(d) 4.4

Answer

Explanation:

Step1: Recall linear - approximation formula

The linear - approximation formula is $L(x)=g(a)+g^{\prime}(a)(x - a)$, where $a$ is the point of tangency, $x$ is the point at which we want to approximate the function value, $g(a)$ is the function value at $a$, and $g^{\prime}(a)$ is the derivative of the function at $a$.

Step2: Identify the values of $a$, $x$, $g(a)$ and $g^{\prime}(a)$

We are given that $a=-1$, $x = - 1.2$, $g(-1)=4$ and $g^{\prime}(-1)=2$.

Step3: Substitute the values into the linear - approximation formula

$L(-1.2)=g(-1)+g^{\prime}(-1)(-1.2-(-1))$. First, simplify the expression inside the parentheses: $-1.2-(-1)=-1.2 + 1=-0.2$. Then, substitute $g(-1)=4$ and $g^{\prime}(-1)=2$ into the formula: $L(-1.2)=4+2\times(-0.2)$. Calculate $2\times(-0.2)=-0.4$. So, $L(-1.2)=4-0.4 = 3.6$.

Answer:

A. 3.6