Let the graph of y = f(x) (solid) and y = g(x) (dashed) be shown below. Suppose P(x) = f(x)g(x). Find P(1) and P'(1).
Understand the Problem
The student is given two graphs, y = f(x) (solid line) and y = g(x) (dashed line). They are asked to calculate P(1) and P'(1) where P(x) = f(x)g(x). To find P(1), we will read the values of f(1) and g(1) from the graph and multiply them. To find P'(1), we will need derive P(x) using the product rule and evaluate at x=1.
Answer
$P(1) = -4$ $P'(1) = -\frac{13}{4}$
Answer for screen readers
$P(1) = -4$
$P'(1) = -\frac{13}{4}$
Steps to Solve
- Find $f(1)$ and $g(1)$ from the graph
From the graph, we read the values of $f(x)$ and $g(x)$ at $x=1$. $f(1) = -1$ $g(1) = 4$
- Calculate $P(1)$
Since $P(x) = f(x)g(x)$, we can find $P(1)$ by substituting $x=1$: $P(1) = f(1)g(1) = (-1)(4) = -4$
- Differentiate $P(x)$ using the product rule
The product rule states that if $P(x) = f(x)g(x)$, then $P'(x) = f'(x)g(x) + f(x)g'(x)$.
- Find $f'(1)$ and $g'(1)$ from the graph
$f'(1)$ is the slope of the tangent to $f(x)$ at $x=1$. Since $f(x)$ is a straight line from $x = -1$ to $x = 5$, we can calculate the slope using two points on this line. The two points are $(-1, 1)$ and $(5, -5)$. $f'(1) = \frac{-5 - 1}{5 - (-1)} = \frac{-6}{6} = -1$ $g'(1)$ is the slope of the tangent to $g(x)$ at $x=1$. We can find the slope using two points on the line formed by $g(x)$. The two points are $(-3, 7)$ and $(1, 4)$. $g'(1) = \frac{4 - 7}{1 - (-3)} = \frac{-3}{4}$
- Calculate $P'(1)$
$P'(1) = f'(1)g(1) + f(1)g'(1) = (-1)(4) + (-1)(-\frac{3}{4}) = -4 + \frac{3}{4} = \frac{-16 + 3}{4} = -\frac{13}{4}$
$P(1) = -4$
$P'(1) = -\frac{13}{4}$
More Information
The product rule is a fundamental concept in calculus and is important for differentiating functions that are products of other functions.
Tips
A common mistake would be to confuse $f(x)$ and $g(x)$ when reading values from the graph, or incorrectly calculating the slopes of the tangent lines at $x=1$. Another mistake is to misapply the product rule. Ensure the correct terms are multiplied together. It is also important to pay attention to the signs of the values.
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