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The diagram below shows the graph of $h'(x)$, the derivative of a cubic function $h(x)$ - Leaving Cert Mathematics - Question 6 - 2021
Question 6
The diagram below shows the graph of $h'(x)$, the derivative of a cubic function $h(x)$. (a) Show that $h'(x) = -2x^2 + 4x + 6$. (b) Use $h'(x)$ to find the ma... show full transcript
Worked Solution & Example Answer:The diagram below shows the graph of $h'(x)$, the derivative of a cubic function $h(x)$ - Leaving Cert Mathematics - Question 6 - 2021
Step 1
Show that $h'(x) = -2x^2 + 4x + 6$
Answer
To prove that , we start by analyzing the graph shown. We can identify key points on the graph, such as the x-intercepts and the vertex.
From the graph, the roots of the polynomial appear to be at and . Thus, we can express in factored form as:
for some constant .
Now we can find using another point from the graph. Observing gives us a y-value from the graph, which we can use to solve for :
- Substitute in the point : 6 = 3k k = 2$$
Thus, we find: To expand this, we calculate: Thus, is validated by this derivation.
Step 2
Use $h'(x)$ to find the maximum positive value of the slope of a tangent to $h(x)$
Answer
To find the maximum positive slope of a tangent to using , we need to determine the maximum value of the function .
We can find the critical points by taking the derivative of : Setting this equal to zero to find critical points:
4x = 4 \ x = 1$$ To determine the nature of this critical point, we can evaluate the second derivative: $$h''(1) = -4(1) + 4 = 0.$$ Since we need to check for maximum through another method, we can look at the values of $h'(x)$ around $x = 1$: Evaluate $h'(0)$ and $h'(2)$: 1. At $x = 0$: $$h'(0) = -2(0^2) + 4(0) + 6 = 6$$ 2. At $x = 2$: $$h'(2) = -2(2^2) + 4(2) + 6 = -8 + 8 + 6 = 6$$ Since both endpoints give us the same maximum positive slope of 6, and $h'(x)$ decreases after $x = 2$, the maximum positive value of the slope of a tangent to $h(x)$ is: **Maximum positive slope = 6**.