Thus

An Advanced Maths Lesson Starter

Diagram

The gradient of \(f(x)\) when \(x=4\) is \(1\).

The area under the curve \(y=f(x)\) between \(x=4\) and \(x=8\) is \(12\) square units.

Which of the following expressions can be evaluated from this information?

Gradients
Gradient to be evaluated Answer
The gradient of \(y=f(x)+3\) at \(x=4\)
The gradient of \(y=-f(x)\) at \(x=4\)
The gradient of \(y=5f(x)\) at \(x=4\)
The gradient of \(y=f(x+2)\) at \(x=4\)
The gradient of \(y=f(x-1)\) at \(x=3\)
The gradient of \(y=f(x+3)\) at \(x=1\)
The gradient of \(y=f(-x)\) at \(x=-4\)
The gradient of \(y=f'(x)\) at \(x=4\)
The gradient of \(y=f''(x)\) at \(x=4\)
The gradient of \(y=f''(x-3)\) at \(x=4\)
Integrals
Integral to be evaluated Answer
\(\displaystyle \int_{4}^{8}\bigl(f(x)+7\bigr)\,dx\)
\(\displaystyle \int_{4}^{8}\bigl(f(x)-5\bigr)\,dx\)
\(\displaystyle \int_{2}^{6}f(x+2)\,dx\)
\(\displaystyle \int_{4}^{8}f(x-2)\,dx\)
\(\displaystyle \int_{4}^{8}-f(x)\,dx\)
\(\displaystyle \int_{4}^{8}f(-x)\,dx\)
\(\displaystyle \int_{4}^{8}5f(x)\,dx\)
\(\displaystyle \int_{-8}^{-4}f(-x)\,dx\)
\(\displaystyle \int_{1}^{2}f(4x)\,dx\)
\(\displaystyle \int_{3}^{7}\bigl(f(x+1)+2\bigr)\,dx\)

 

Answers

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