Find the Local Maxima and Minima f(x)=5x^2+4x-6

Problem

ƒ(x)=5*x2+4*x−6

Solution

1. Find the derivative of the function to determine the slope of the tangent line.

d(ƒ(x))/d(x)=10*x+4

2. Set the derivative to zero to find the critical points where the slope is horizontal.

10*x+4=0

3. Solve for x by isolating the variable.

10*x=−4

x=−4/10

x=−0.4

4. Find the second derivative to apply the Second Derivative Test and determine the concavity.

d2(ƒ(x))/(d(x)2)=10

5. Analyze the concavity at the critical point. Since the second derivative is 10 which is greater than 0 the function is concave up, indicating a local minimum.

10>0⇒local minimum at *x=−0.4

6. Calculate the y-coordinate by substituting the critical value back into the original function.

ƒ*(−0.4)=5*(−0.4)2+4*(−0.4)−6

ƒ*(−0.4)=5*(0.16)−1.6−6

ƒ*(−0.4)=0.8−1.6−6

ƒ*(−0.4)=−6.8

Final Answer

Local Minimum: *(−0.4,−6.8), Local Maxima: None

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