Question

1. (2 points) use the figure to find the values of a, b, c, and d about f at p, where f(a)=b and f(c)=d

a =

b =

c =

d =

1. (5 points) let f(x)=2x^{2}-4x + 6. find the derivative f(x) by using the definition of the derivative of f as a limit of a difference quotient, and use algebraic calculations to compute the limit.

Explanation:

Step1: Identify the value of \(a\)

\(a\) is the \(x\) - coordinate of the point \(P\). From the figure, the point \(P\) has coordinates \((4,1)\), so \(a = 4\).

Step2: Identify the value of \(b\)

Since \(f(a)=b\) and \(a = 4\), \(b\) is the \(y\) - coordinate of the point on the function \(y = f(x)\) at \(x=a\). For \(x = 4\), from the point \(P(4,1)\) on \(y = f(x)\), \(b = 1\).

Step3: Identify the value of \(c\)

\(c\) is the \(x\) - coordinate at which we are finding the derivative. The derivative \(f^{\prime}(c)\) is evaluated at the \(x\) - coordinate of the point of tangency. The point of tangency is \(P\) with \(x\) - coordinate \(4\), so \(c = 4\).

Step4: Find the value of \(d\)

\(d=f^{\prime}(c)\) and \(c = 4\). The slope of the tangent line at the point \((x_1,y_1)=(4,1)\) and another point \((x_2,y_2)=(2,5)\) on the tangent - line can be found using the slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\). So \(d=\frac{5 - 1}{2 - 4}=\frac{4}{-2}=-2\).

Answer:

\(a = 4\)
\(b = 1\)
\(c = 4\)
\(d=-2\)